Maharashtra Board Class 9 Science Chapter 2 Work and Energy Solutions

Std 9 Science Chapter 2 Work And Energy Question Answer Maharashtra Board

Class 9 Science Chapter 2 Work And Energy Question Answer Maharashtra Board

1. Write Detailed Answers?

Question 1.a. Explain the difference between potential energy and kinetic energy.
Answer:

🎯 Exam Tip: When differentiating, ensure clear contrasting points for definition, formula, and examples for full marks.

Question 1.b. Derive the formula for the kinetic energy of an object of mass m, moving with velocity v.
Answer:Suppose a stationary object of mass 'm' moves because of an applied force. Let 'u' be its initial velocity (here u = 0). Let the applied force be 'F'. This generates an acceleration a in the object, and after time T, the velocity of the object becomes equal to 'v'. The displacement during this time is s. The work done on the object is \[W = F \times s \quad (1)\] Using Newton's 2nd law of motion, \[F = ma \quad (2)\] Using Newton's 2nd equation of motion \[s = ut + \frac{1}{2}at^2\] However, as initial velocity is zero, u = 0 \[s = 0 + \frac{1}{2}at^2\] \[\therefore s = \frac{1}{2}at^2 \quad (3)\] Using equations (2) and (3) in (1) \[W = ma \times \frac{1}{2}at^2\] \[\therefore W = \frac{1}{2}m \times (at)^2 \quad (4)\] Using Newton's first equation of motion \[v = u + at\] \[v = 0 + at\] \[\therefore v = at\] \[\therefore v^2 = a^2t^2 = (at)^2 \quad (5)\] using equations (4) and (5) \[\therefore W = \frac{1}{2}mv^2\] The kinetic energy gained by an object is the amount of work done on the object. \[\therefore K.E = W\] \[\therefore K.E = \frac{1}{2}mv^2\]In simple words: This derivation shows how work done on an object to change its velocity is stored as kinetic energy, linking force, displacement, and motion equations.

🎯 Exam Tip: Remember to clearly state the assumptions (e.g., initial velocity, applied force) and the laws of motion used at each step of the derivation.

Question 1.c. Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.
Answer:Let us look at the kinetic and potential energies of an object of mass (m), falling freely from height (h), when the object is at different heights.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक स्वतंत्र रूप से गिरती हुई वस्तु के तीन बिंदुओं पर उसकी ऊर्जाओं को दर्शाता है। बिंदु A वस्तु की प्रारंभिक स्थिति है जो जमीन से 'h' ऊंचाई पर है। बिंदु B एक मध्यवर्ती स्थिति है जो A से 'x' दूरी पर है (जमीन से ऊंचाई 'h-x')। बिंदु C जमीन पर अंतिम स्थिति है। इसका उपयोग वस्तु की कुल यांत्रिक ऊर्जा को तीनों बिंदुओं पर समान साबित करने के लिए किया जाता है। As shown in the figure, the point A is at a height (h) from the ground. Let the point B be at a distance x, vertically below A. Let the point C be on the ground directly below A and B. Let us calculate the energies of the object at A, B and C. (1) Let the velocity of the object be \(v_B\) when it reaches point B, having fallen through a distance x. \[u = 0, s = x, a = g\] \[v^2 = u^2 + 2as\] \[\therefore v_B^2 = 0 + 2gx\] \[\therefore v_B^2 = 2gx\] \[K.E = \frac{1}{2}mv^2 = \frac{1}{2}m(2gx)\] \[\therefore K.E = mgx\] Height of the object when at B = h-x \[\therefore P.E = mg (h-x)\] \[\therefore P.E = mgh - mgx\] \[\therefore \text{Total Energy T.E.} = K.E + P.E\] \[\implies \text{T.E.} = mgx + mgh - mgx\] \[\therefore \text{T.E.} = mgh \quad \text{...(i)}\] (2) When the object is stationary at A, its initial velocity is u = 0 \[\therefore K.E = \frac{1}{2}mu^2\] \[\therefore K.E = 0\] \[P.E = mgh\] \[\therefore \text{Total energy} = K.E + P.E\] \[\implies \text{Total energy} = 0 + mgh\] \[\therefore \text{Total Energy} = mgh. \quad \text{...(ii)}\] (3) Let the velocity of the object be \(v_C\) when it reaches the ground, near point C. \[u = 0, s = h, a = g\] \[v^2 = u^2 + 2as\] \[\therefore v_C^2 = 0 + 2gh\] \[K.E = \frac{1}{2}mv_C^2 = \frac{1}{2}m(2gh)\] \[\therefore K.E = mgh\] The height of the object from the ground at point C is h = 0 \[\therefore P.E = mgh = 0\] \[\therefore \text{T.E. - K.E + P.E}\] \[\therefore \text{T.E.} = mgh \quad \text{...(iii)}\] From equations (i) and (iii) we see that the total potential energy of the object at its initial position is the same as the kinetic energy at the ground.In simple words: This proof demonstrates that as an object falls, its potential energy converts entirely into kinetic energy, illustrating the principle of conservation of mechanical energy.

🎯 Exam Tip: Ensure you clearly define the energy at each point (initial, intermediate, final) and show how the sum of kinetic and potential energy remains constant, assuming no energy loss due to air resistance.

Question 1.d. Determine the amount of work done when an object is displaced at an angle of 30 degrees with respect to the direction of the applied force.
Answer:When an object is displaced by displacement 's' and by applying force 'F' at an 'angle' \(30^\circ\), work done can be given as \[W = Fs \cos \theta\] \[\implies W = Fs \cos 30^\circ \quad (\theta = 30^\circ)\] \[\implies W = Fs \left(\frac{\sqrt{3}}{2}\right) \quad \left(\cos 30^\circ = \frac{\sqrt{3}}{2}\right)\]In simple words: When force and displacement are not in the same direction, only the component of force parallel to the displacement does work.

🎯 Exam Tip: Always remember that work is a scalar product of force and displacement. The angle between the two vectors is crucial for calculating work done using the formula \(W = Fs \cos \theta\).

Question 1.e. If an object has 0 momenta, does it have kinetic energy? Explain your answer.
Answer:

• No, it does not have kinetic energy if it does not have momentum.
• Momentum is the product of mass and velocity. If it is zero, it implies that \(v = 0\) (since mass can never be zero).
• Now \(K.E = \frac{1}{2}mv^2\), So if \(v = 0\) then K.E also will be zero.
• Thus, if an object has no momentum then it cannot possess kinetic energy.

🎯 Exam Tip: Clearly define both momentum (\(p = mv\)) and kinetic energy (\(K.E. = \frac{1}{2}mv^2\)) to show their direct dependence on velocity.

Question 1.f. Why is the work done on an object moving with uniform circular motion zero?
Answer:

• In uniform circular motion, the force acting on an object is along the radius of the circle.
• Its displacement is along the tangent to the circle. Thus, they are perpendicular to each other.
• Hence \(\theta = 90^\circ\) and \(\cos 90^\circ = 0\)
• \(\therefore W = Fs \cos 0 = 0\)

🎯 Exam Tip: This is a common conceptual question. Emphasize the \(90^\circ\) angle between force and displacement and the value of \(\cos 90^\circ\) for a clear explanation.

2. Choose One Or More Correct Alternatives.

Question 2.a. For work to be performed, energy must be ....
(i) transferred from one place to another
(ii) concentrated
(iii) transformed from one type to another
(iv) destroyed
Answer: (iii) transformed from one type to another
In simple words: Work involves a change, and in physics, this change often means energy changing from one form to another.

🎯 Exam Tip: Understand that energy transformation is the fundamental principle behind work. Memorizing examples of different energy conversions is helpful.

Question 2.b. Joule is the unit of ...
(i) force
(ii) work
(iii) power
(iv) energy
Answer: (ii) work
In simple words: Joule is the standard unit for both work and energy, representing the amount of energy transferred when a force of one Newton moves an object one meter.

🎯 Exam Tip: Pay attention to SI units. Joule is for work and energy, Newton for force, Watt for power.

Question 2.c. Which of the forces involved in dragging a heavy object on a smooth, horizontal surface, have the same magnitude?
(i) the horizontal applied force
(ii) gravitational force
(iii) reaction force in vertical direction
(iv) force of friction
Answer: (Answer not provided in the source text.)
In simple words: On a smooth horizontal surface, if there is no vertical acceleration, the downward gravitational force and the upward normal (reaction) force are equal in magnitude.

🎯 Exam Tip: Remember Newton's Third Law (action-reaction pairs) and consider forces in equilibrium along different axes. For a horizontal surface, vertical forces often balance out.

Question 2.d. Power is a measure of the ......
(i) the rapidity with which work is done
(ii) amount of energy required to perform the work
(iii) The slowness with which work is performed
(iv) length of time
Answer: (i) the rapidity with which work is done
In simple words: Power tells us how fast work is done or how quickly energy is transferred.

🎯 Exam Tip: Remember the formula for power, \(P = W/t\), to recall that it's about the rate of doing work, not just the amount.

Question 2.e. While dragging or lifting an object, negative work is done by
(i) the applied force
(ii) gravitational force
(iii) frictional force
(iv) reaction force
Answer: (Answer not provided in the source text.)
In simple words: Negative work occurs when the force applied is in the opposite direction to the object's displacement.

🎯 Exam Tip: To identify negative work, analyze the direction of the specific force relative to the direction of motion. Forces like friction (when dragging) or gravity (when lifting) often do negative work.

3. Rewrite The Following Sentences Using A Proper Alternative.

Question 3.a. The potential energy of your body is least when you are ....
(i) sitting on a chair
(ii) sitting on the ground
(iii) sleeping on the ground
(iv) standing on the ground
Answer: (iii) sleeping on the ground
In simple words: Potential energy is related to height, so the lowest possible position for your body will result in the least potential energy.

🎯 Exam Tip: Remember that potential energy (\(P.E. = mgh\)) is relative to a chosen reference point. The lower the height (\(h\)), the lower the potential energy.

Question 3.b. The total energy of an object falling freely towards the ground ...
(i) decreases
(ii) remains unchanged
(iii) increases
(iv) increases in the beginning and then decreases
Answer: (iii) increases
In simple words: According to the principle of conservation of energy, the total mechanical energy (potential + kinetic) of an object falling freely in a vacuum remains constant, though individual forms may transform.

🎯 Exam Tip: The law of conservation of energy states that in an isolated system, total energy is conserved. While potential energy converts to kinetic energy, their sum stays constant. This question's provided answer (iii) 'increases' is generally considered incorrect in physics for total energy in a free fall.

Question 3.c. If we increase the velocity of a car moving on a flat surface to four times its original speed, its potential energy ....
(i) will be twice its original energy
(ii) will not change
(iii) will be 4 times its original energy
(iv) will be 16 times its original energy.
Answer: (ii) will not change
In simple words: Potential energy depends on height and position, not on the horizontal speed of an object.

🎯 Exam Tip: Differentiate between kinetic energy (\(\frac{1}{2}mv^2\)) which depends on speed, and potential energy (\(mgh\)) which depends on height. Changes in horizontal velocity do not affect potential energy on a flat surface.

Question 3.d. The work done on an object does not depend on ....
(i) displacement
(ii) applied force
(iii) initial velocity of the object
(iv) the angle between force and displacement.
Answer: (iii) initial velocity of the object
In simple words: Work done is determined by the force, displacement, and the angle between them, not by the object's initial speed.

🎯 Exam Tip: Recall the work formula \(W = Fs \cos \theta\). This formula clearly shows the factors influencing work done. Initial velocity affects kinetic energy change, but not directly the work done calculation itself.

4. Study The Following Activity And Answer The Questions.
1. Take two aluminium channels of different lengths.
2. Place the lower ends of the channels on the floor and hold their upper ends at the same height.
3. Now take two balls of the same size and weight and release them from the top end of the channels. They will roll down and cover the same distance.

Questions

Question 4.1. At the moment of releasing the balls, which energy do the balls have?
Answer: At the moment of releasing the ball they possess Potential energy as they are at a height above the ground.
In simple words: When the balls are held at a height, they store potential energy due to their elevated position.

🎯 Exam Tip: Potential energy is stored energy due to position or state. Objects at rest at a height primarily possess potential energy.

Question 4.2. As the balls roll down which energy is converted into which other form of energy?
Answer: As the balls roll down, the Potential energy is converted into Kinetic energy since they are now in motion.
In simple words: As the balls move downwards, their stored potential energy is transformed into energy of motion, known as kinetic energy.

🎯 Exam Tip: This illustrates energy transformation – potential to kinetic – a key concept in the conservation of mechanical energy.

Question 4.3. Why do the balls cover the same distance on rolling down?
Answer: Since they have been released from the same height, they will cover the same distance.
In simple words: If released from the same height, and assuming minimal energy loss, the balls will have the same energy conversion, leading to similar motion and distance covered.

🎯 Exam Tip: This outcome is related to the initial potential energy being the same, which leads to the same amount of kinetic energy and thus similar motion patterns.

Question 4.4. What is the form of the eventual total energy of the balls?
Answer: The eventual form of the total energy of the balls is "Mechanical Energy" i.e, a combination of Potential energy and Kinetic energy
In simple words: The total energy of the balls remains mechanical energy, which is the sum of their potential and kinetic energies throughout their motion.

🎯 Exam Tip: Mechanical energy is the sum of potential and kinetic energy. In ideal scenarios (no friction), total mechanical energy is conserved.

Question 4.5. Which law related to energy does the above activity demonstrate? Explain.
Answer: The above activity demonstrates the “Law of Conservation of Energy"
In simple words: This activity shows that energy cannot be created or destroyed, only transformed from one form to another, specifically potential to kinetic.

🎯 Exam Tip: Be ready to explain the Law of Conservation of Energy with examples of energy transformations, emphasizing that the total energy in a closed system remains constant.

5. Solve The Following Examples.

Question 5.a. An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m? (Ans : 1224.5 kg)
Answer:Given:
Power (P) = 2 kW = 2000 W
Height (h) = 10 m
Time (t) = 1 min = 60 s
Acceleration due to gravity (g) = 9.8 m/s²
To Find:
Mass of water (m)= ?
Formula:
\[P = \frac{mgh}{t}\]
Solution:
\[P = \frac{mgh}{t}\]
\[\implies m = \frac{Pt}{gh}\]
\[\implies m = \frac{2000 \times 60}{9.8 \times 10}\]
\[\implies m = \frac{120000}{98}\]
\[\implies m = 1224.5 \text{ kg}\] Water lifted by the pump is 1224.5 kg
In simple words: This problem uses the power formula, which relates power, mass, gravity, height, and time, to calculate how much water an electric pump can lift.

🎯 Exam Tip: Always convert units to SI (kW to W, minutes to seconds) before applying formulas. Pay attention to the correct rearrangement of the power formula to solve for the unknown variable.

Question 5.b. If the energy of a ball falling from a height of 10 metres is reduced by 40%, how high will it rebound? (Ans : 6 m)
Answer:Given: Initial height (h₁) = 10m
Let Initial (P.E₁) = 100
Final (P.E₂) = 100 – 40
= 60
To Find:
Final height (h2) = ?
Formula:
P.E. = mgh
Solution:
\[P.E₁ = mgh₁ \quad \text{(i)}\] \[P.E₂ = mgh₂ \quad \text{(ii)}\] Dividing (ii) by (i)
\[\implies \frac{P.E₂}{P.E₁} = \frac{mgh₂}{mgh₁}\]
\[\implies \frac{60}{100} = \frac{h₂}{10}\]
\[\implies h₂ = \frac{60}{100} \times 10\]
\[\implies h₂ = 6 \text{ m}\] The ball will rebound by 6 m.
In simple words: This problem calculates the rebound height of a ball by finding its final potential energy after a percentage loss, directly relating energy to height.

🎯 Exam Tip: When dealing with percentage reductions in energy, it's often easiest to work with ratios or assume an initial arbitrary value (like 100 units) to find the final value before converting back to physical units.

Question 5.d. The velocity of a car increase from 54 km/hr to 72 km/hr. How much is the work done if the mass of the car is 1500 kg? (Ans. : 131250 J)
Answer:Given: Mass (m) = 1500 kg
Initial velocity (u) = 54 km/hr
\[\implies u = 54 \times \frac{5}{18}\]
\[\implies u = 15 \text{ m/s}\] Final velocity (v) = 72 km/hr
\[\implies v = 72 \times \frac{5}{18}\]
\[\implies v = 20 \text{m/s}\] To Find:
Work done to increase the velocity = ?
Formula:
Work done to increase velocity = Change in K.E
\[\text{Change in K.E} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\] Solution:
\[\text{Change in K.E} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\]
\[\implies \text{Change in K.E} = \frac{1}{2}m (v^2 - u^2)\]
\[\implies \text{Change in K.E} = \frac{1}{2} \times 1500 [(20)^2 - (15)^2]\]
\[\implies \text{Change in K.E} = 750 \times [400-225]\]
\[\implies \text{Change in K.E} = 750 \times 175\]
\[\implies \text{Change in K.E} = 131250 \text{ J}\] Work done to increase the velocity = 131250 J
In simple words: The work-energy theorem states that the net work done on an object equals its change in kinetic energy, which is used here to calculate work from velocity change.

🎯 Exam Tip: Crucially, convert all velocities from km/hr to m/s before calculation. Remember that work done in changing velocity is equal to the change in kinetic energy.

Question 5.e. Ravi applied a force of 10 N and moved a book 30 cm in the direction of the force. How much was the work done by Ravi? (Ans: 3 J)
Answer:Given:
Force (F) = 10 N
\(\theta = 0^\circ\), (Since force and displacement are in same direction)
Displacement (s) = 30 cm = 30/100 m
To Find:
Work (W) = ?
Formula:
W = Fs \(\cos \theta\)
Solution:
W = Fs \(\cos \theta\)
\[\implies W = 10 \times \frac{30}{100} \times \cos 0 \quad (\cos 0 = 1)\]
\[\implies W = 3 \text{ J}\] The work done by Ravi is 3J
In simple words: When force and displacement are in the same direction, work done is simply the product of force and distance moved.

🎯 Exam Tip: Always convert displacement to meters (SI unit) before calculating work. For forces in the direction of motion, \(\cos \theta = 1\).

Class 9 Science Chapter 1 Laws Of Motion Intext Questions And Answers

Question 1. What are different types of forces? Give examples.
Answer:Forces are of two types.
(i) Contact force e.g.: Mechanical force, frictional force, muscular force
(ii) Non-contact force e.g.: gravitational force, magnetic force, electrostatic force
In simple words: Forces can be categorized into contact forces (requiring physical touch) and non-contact forces (acting at a distance).

🎯 Exam Tip: When asked for types of forces, always provide a clear definition for each type and illustrate with relevant, distinct examples.

Question 2. Monashee wants to displace a wooden block from point A to point B along the surface of a table as shown. She has used force F for the purpose.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक व्यक्ति (मोनाशी) को मेज पर एक लकड़ी के ब्लॉक को बिंदु A से बिंदु B तक खिसकाते हुए दिखाता है। व्यक्ति ब्लॉक पर बल 'F' लगा रहा है और ब्लॉक मेज की सतह पर विस्थापित हो रहा है।
(a) Has all the energy she spent been used to produce an acceleration in the block
(b) Which forces have been overcome using that energy?
Answer:
(a) Only part of the energy applied by Minakshee is used in accelerating the block.
(b) Force of friction has been overcome using the energy.
In simple words: Not all energy applied goes into acceleration; some is always used to overcome opposing forces like friction.

🎯 Exam Tip: Understand that in real-world scenarios, energy is often "lost" to non-conservative forces like friction or air resistance, meaning not all applied energy contributes to desired motion or acceleration.

Question 3. Mention the type of energy used in the following examples.
(i) Stretched rubber string.
(ii) Fast-moving car.
(iii) The whistling of a cooker due to steam.
(iv) A fan running on electricity.
(v) Drawing out pieces of iron from garbage, using a magnet.
(vi) Breaking of a glass window pane because of a loud noise.
(vii) The drackers exploded in Diwali.
Answer:
(i) Potential energy
(ii) Kinetic energy
(iii) Sound energy
(iv) Electrical energy
(v) Magnetic energy
(vi) Sound energy
(vii) Sound energy, light energy and heat energy
In simple words: Energy exists in various forms, such as potential (stored), kinetic (motion), sound, electrical, magnetic, light, and heat, often transforming between them.

🎯 Exam Tip: Be able to identify and categorize different forms of energy based on given scenarios. This demonstrates understanding of energy transformations.

Question 4. Study the pictures given below and answer the questions:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र चार अलग-अलग गतिविधियों को दर्शाता है: एक लड़की पढ़ रही है, एक लड़का गेंद से खेल रहा है, एक लड़की टीवी देख रही है, और एक व्यक्ति अनाज का बोरा उठा रहा है। इसका उद्देश्य यह विश्लेषण करना है कि इनमें से किन स्थितियों में कार्य किया गया है।
(a) In which of the pictures above has work been done?
(b) From scientific point of view, when do we say that no work was done?
Answer:
(a) Girl studying : No work done
Boy playing with ball: Work is done
Girl watching T.V.: No work done
Person lifting sack of grains : Work is done
(b) No work is said to be done when force is applied but there is no displacement.
In simple words: Work in physics requires both a force and displacement in the direction of that force. If there's no movement or the movement is perpendicular to the force, no work is done.

🎯 Exam Tip: To determine if work is done, identify the applied force and check for displacement in the direction of that force. Understanding the formula \(W = Fs \cos \theta\) is key.

Question 5. Make two pendulums of the same length with the help of thread and two nuts. Tie another thread in the horizontal position.
Tie the two pendulums to the horizontal thread in such a way that they will not hit each other while swinging. Now swing one of the pendulums and observe. What do you see?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो युग्मित दोलक (coupled oscillators) दिखाता है। एक क्षैतिज धागे से दो समान लंबाई के दोलक (धागे और नट से बने) लटके हुए हैं। यह व्यवस्था ऊर्जा के हस्तांतरण को प्रदर्शित करती है जब एक दोलक को झुलाया जाता है।
Answer:You will see that as the speed of oscillation of the pendulum slowly decreases, the second pendulum which was initially stationary, begins to swing. Thus, one pendulum transfers its energy to the other.
In simple words: This experiment demonstrates sympathetic oscillation and energy transfer, where the vibrations of one pendulum cause a nearby, initially stationary pendulum to start swinging.

🎯 Exam Tip: This activity illustrates resonant energy transfer. Understand that energy from one vibrating system can be efficiently transferred to another with a similar natural frequency.

Question 6. Ajay and Atul have been asked to determine the potential energy of a ball of mass m kept on a table as shown in the figure. What answers will they get? Will they be different? What do you conclude from this?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक मेज पर रखी 'm' द्रव्यमान की गेंद की संभावित ऊर्जा का आकलन करने के लिए अजय और अतुल की स्थितियों को दर्शाता है। अजय गेंद की ऊंचाई 'h1' को मेज से माप रहा है, जबकि अतुल उसे जमीन से 'h2' ऊंचाई पर माप रहा है, यह दर्शाता है कि संभावित ऊर्जा संदर्भ बिंदु पर निर्भर करती है।
Answer:

• According to Ajay P.E₁ = mgh₁ and according to Atul P.E₂ = mgh₂.
• Yes, the answer will be different as the two heights are different.
• Potential energy is relative.

🎯 Exam Tip: Emphasize that potential energy is not an absolute quantity. Always specify the reference level (e.g., ground, table surface) when calculating potential energy.

Question 7. Discuss the directions of force and of displacement in each of the following cases.
(i) Pushing a stalled vehicle.
(ii) Catching the ball which your friend has thrown towards you.
(iii) Tying a stone to one end of a string and swinging it round and round by the other end of the string.
(iv) Walking up and down a staircase; climbing a tree.
(v) Stopping a moving car by applying brakes.
Answer:
(i) Force and displacement are in the same direction.
(ii) Force and displacement are in the opposite direction.
(iii) Force and displacement are perpendicular to each other.
(iv) Force and displacement are in the opposite direction.
(v) Force and displacement are in the opposite direction.
In simple words: The relationship between force and displacement direction dictates the type of work done: same direction (positive), opposite (negative), or perpendicular (zero).

🎯 Exam Tip: For each scenario, visualize the force acting and the resulting motion. Use the angle \(\theta\) between force and displacement to determine if work is positive (\(\theta < 90^\circ\)), negative (\(\theta > 90^\circ\)), or zero (\(\theta = 90^\circ\)).

Question 8.
(A) An arrow is released from a stretched bow.
(B) Water kept at a high flows through a pipe into the tap below.
(C) A compressed spring is released.
(a) Which words describe the state of the object in the above examples?
(b) Where did the energy required to cause the motion of the objects come from?
(c) If the objects were not brought in those states, would they have moved?
Answer:
(a) Words such as stretched bow, water kept at a height and compressed spring describe the state of the objects.
(b) The energy required for the objects came from its specific state or motion in the form of potential energy.
(c) No, if the objects were not brought in those states, they would have not moved.
In simple words: This question explores different forms of potential energy - elastic potential energy in a stretched bow or compressed spring, and gravitational potential energy in water at a height. These objects store energy due to their state or position, which is then converted into kinetic energy to cause motion. Without this stored energy, they wouldn't move on their own.

🎯 Exam Tip: Understanding the different types of potential energy (gravitational, elastic) and their role in initiating motion is crucial for conceptual clarity and scoring well.

Question 9.
Study the activity and answer the following questions.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र तीन अलग-अलग स्थितियों को दर्शाता है जहाँ एक कप को धागे और नट की मदद से एक मेज पर रखा गया है। चित्र A में, धागा मेज के किनारे से नीचे लटका हुआ है और नट कप को खींच रहा है जिससे विस्थापन होता है। चित्र B में, एक शासक का उपयोग करके कप को विपरीत दिशा में धकेला जा रहा है, जिससे नकारात्मक कार्य होता है। चित्र C में, दो बराबर और विपरीत बल कप पर लग रहे हैं, जिससे कोई विस्थापन नहीं होता है और कार्य शून्य होता है।
Positive, negative and zero work
(a) Figure A - Why does the cup get pulled?
(b) Figure B - What is the relation between the displacement of the cup and the force applied through the ruler?
(c) In Figure C-Why doesn't the cup get displaced?
(d) What is the type of work done in figures A, B and C?
(e) In the three actions above, what is the relationship between the applied force and the displacement?
Answer:
(a) The cup gets pulled as the force of the nut and the displacement of the cup is in the same direction.
(b) The displacement of the cup and the force applied through the ruler is in the opposite direction.
(c) The cup does not get displaced as two equal forces are working in opposite directions.
(d) The work done in figure A is positive, figure B is negative and in figure C is zero.
(e) In figure A the applied force and the displacement is in the same direction, in figure B the applied force and the displacement is in the opposite direction and in figure C the applied force and displacement is perpendicular to each other.
In simple words: This activity illustrates positive, negative, and zero work based on the direction of force relative to displacement. Positive work occurs when force and displacement are in the same direction, negative work when they are opposite, and zero work when they are perpendicular or when there is no displacement despite force being applied.

🎯 Exam Tip: This question directly tests the understanding of the conditions for positive, negative, and zero work. Clear conceptual understanding of the angle between force and displacement is key.

Question 10.
From the following activities find out whether work is positive, negative or zero.
Give reasons for your answers.
(a) A boy is swimming in a pond.
(b) A coolie is standing with a load on his head.
(c) Stopping a moving car by applying brakes.
(d) Catching the ball which you friend has thrown towards you.
Answer:
(a) A boy is swimming in a pond: The work done is positive because the direction of applied force and displacement are the same.
(b) A coolie is standing with a load on his head: The work done is zero because the applied force does not cause any displacement.
(c) Stopping a moving car by applying brakes: The work done is negative because the force applied by the brakes acts in a direction opposite to the direction of motion of car.
(d) Catching the ball which you friend has thrown towards you : Negative work because the force required to stop the ball, acts opposite to the displacement of the ball.
In simple words: Work can be positive if force helps motion, negative if force opposes motion, or zero if force is perpendicular to motion or causes no movement. Examples like swimming (positive), holding a load (zero), braking a car (negative), and catching a ball (negative) illustrate these principles.

🎯 Exam Tip: Being able to identify and explain positive, negative, and zero work in everyday scenarios demonstrates a strong grasp of the work-energy principle.

Question 11.
(a) Can your father climb stairs as fast as you can?
(b) Will you fill the overhead water tank with the help of a bucket or an electrical motor?
(c) Suppose Rajashree, Yash and Ranjeet have to reach the top of a small hill. Rajashree went by car. Yash went cycling while Ranjeet went walking. If all of them choose the same path, who will reach first and who will reach last? (Think before you answer.
Answer:
(a) No, father takes more time to climb stairs.
(b) Overhead water tank can be filled with the help of one electric motor rather than filling it with bucket.
(c) Rajashree will reach first, followed by Yash and Ranjeet will reach last because car moves faster than a cycle and a person walking.
In simple words: This question focuses on the concept of power and efficiency. Younger individuals or machines like electric motors can perform work faster (have more power). Vehicles like cars are faster than cycles, which are faster than walking, illustrating varying rates of work or power.

🎯 Exam Tip: This question highlights the practical application of power-how quickly work can be done. Higher power means less time taken for the same amount of work.

Class 9 Science Chapter 1 Laws of Motion Additional Important Questions And Answers

Question 1.
Forces are of ......................... types.
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (a) 2
In simple words: Forces are broadly classified into two main types: contact forces, which require physical touch, and non-contact forces, which act from a distance.

🎯 Exam Tip: Knowing the fundamental classification of forces is a basic but important concept. Make sure to remember the two main types.

Question 2.
Example of Contact force is ......................... .
(a) Gravitational Force
(b) Magnetic Force
(c) Electrostatic Force
(d) Muscular Force
Answer: (d) Muscular Force
In simple words: A contact force is one that acts only when two objects are physically touching each other. Muscular force, like pushing or pulling, requires direct contact.

🎯 Exam Tip: Distinguish between contact and non-contact forces by recalling examples for each. Muscular force is a classic example of contact force.

Question 3.
Example of Non-contact force is .........................
(a) Mechanical Force
(b) Frictional Force
(c) Muscular Force
(d) Electrostatic Force
Answer: (d) Electrostatic force
In simple words: A non-contact force is one that acts on an object without physical contact. Electrostatic force, like the attraction between charged particles, can act over a distance.

🎯 Exam Tip: Non-contact forces (gravitational, magnetic, electrostatic) are those that can exert influence without touching, which is a key characteristic to remember.

Question 4.
Work is said to be done on a body when a ......................... is applied on object causes displacement of the object.
(a) Direction
(b) Area
(c) Volume
(d) Force
Answer: (d) force
In simple words: Work is done when a force acts on an object and causes it to move a certain distance in the direction of the force. Both force and displacement are essential for work to be performed.

🎯 Exam Tip: The definition of work is fundamental: force applied over a distance. Remember that without either, no work is done.

Question 5.
W = .........................
(a) mgh
(b) mdh
(c) mv²
(d) mfe
Answer: (a) mgh
In simple words: The formula W = mgh represents the work done against gravity or the potential energy of an object at a certain height, where 'm' is mass, 'g' is acceleration due to gravity, and 'h' is height.

🎯 Exam Tip: Be familiar with common physics formulas. \( W = mgh \) is specifically for work done against gravity or for gravitational potential energy.

Question 6.
The energy stored in the dry cell is in of ......................... energy.
(a) Light
(b) Chemical
(c) Solar
(d) Kinetic
Answer: (b) chemical
In simple words: Dry cells, or batteries, store energy in chemical compounds. This chemical energy is converted into electrical energy when the cell is used.

🎯 Exam Tip: Identify different forms of energy and where they are stored. Chemical energy is common in fuels and batteries.

Question 7.
The work done is zero if there is no .........................
(a) Direction
(b) Displacement
(c) Mass
(d) Angle
Answer: (b) displacement
In simple words: According to the definition of work (Work = Force × Displacement × cosθ), if there is no displacement, then no work is done, regardless of the force applied.

🎯 Exam Tip: Remember the conditions for zero work: either no force, no displacement, or force perpendicular to displacement.

Question 8.
Flowing water has ......................... energy.
(a) Potential
(b) Chemical
(c) Solar
(d) Kinetic
Answer: (d) kinetic
In simple words: Flowing water is in motion, and any object or substance in motion possesses kinetic energy.

🎯 Exam Tip: Kinetic energy is the energy of motion. Any moving object, be it water, a car, or a person, has kinetic energy.

Question 9.
By stretching the rubber strings of a we store ......................... energy in it.
(a) Potential
(b) Chemical
(c) Electric
(d) Kinetic
Answer: (a) potential
In simple words: When a rubber string is stretched, energy is stored in it due to its change in shape. This stored energy is a form of elastic potential energy.

🎯 Exam Tip: Elastic potential energy is stored when an object is deformed (stretched, compressed) and has the capacity to return to its original shape.

Question 10.
......................... is the unit of force.
(a) Both B and C
(b) Newton
(c) Dyne
(d) Volts
Answer: (a) Both B and C
In simple words: Both Newton (SI unit) and Dyne (CGS unit) are standard units used to measure force. Volts measure electric potential.

🎯 Exam Tip: Know the SI and CGS units for fundamental quantities like force. Newton is the SI unit, and dyne is the CGS unit.

Question 11.
For a freely falling body, kinetic energy is ......................... at the ground level.
(a) Maximum
(b) Minimum
(c) Neutral
(d) Reversed
Answer: (a) Maximum
In simple words: As a body falls freely, its potential energy converts into kinetic energy. At ground level, just before impact, all its initial potential energy is converted to kinetic energy, making its kinetic energy maximum and velocity highest.

🎯 Exam Tip: The principle of conservation of energy dictates that as an object falls, its potential energy decreases while its kinetic energy increases, reaching maximum kinetic energy just before hitting the ground.

Question 12.
Energy can neither be ......................... nor .........................
(a) Destroyed
(b) Created
(c) Saved
(d) Both A and B
Answer: (d) Both A and B
In simple words: This statement refers to the Law of Conservation of Energy, which states that energy cannot be created from nothing or destroyed into nothing, but rather it can only be transformed from one form to another.

🎯 Exam Tip: The law of conservation of energy is a core principle in physics. Remember that energy only changes forms, it's not made or unmade.

Question 13.
Work and ......................... have the same unit.
(a) Energy
(b) Electricity
(c) Force
(d) Both B and C
Answer: (a) Energy
In simple words: Work and energy are closely related concepts, both representing the capacity to do something or the result of something being done. Therefore, they share the same SI unit, which is the Joule.

🎯 Exam Tip: Work and energy are interchangeable in terms of their units because work is the transfer of energy. Both are measured in Joules.

Question 14.
S.I. unit of energy is .........................
(a) Joule
(b) Ergs
(c) m/s²
(d) Both A and B
Answer: (a) Joule
In simple words: The standard international (SI) unit for measuring energy is the Joule (J). Ergs are a unit of energy in the CGS system, and m/s² is a unit of acceleration.

🎯 Exam Tip: Always remember the SI units for basic physical quantities. Joule is the standard for energy and work.

Question 15.
Work is the product of .........................
(a) force and distance
(b) displacement and velocity
(c) kinetic and potential energy
(d) force and displacement
Answer: (d) force and displacement
In simple words: Work is precisely defined as the product of the force applied to an object and the displacement of the object in the direction of that force.

🎯 Exam Tip: The most fundamental definition of work is force multiplied by displacement in the direction of the force. Be precise with terms like 'distance' vs. 'displacement'.

Question 16.
S.I. unit of work is .........................
(a) dyne
(b) newton-meter or erg
(c) N/m2 or joule
(d) newton-meter or joule
Answer: (d) newton-meter or joule
In simple words: The SI unit for work is the Joule (J). A Joule is equivalent to one Newton-meter (N·m), which is the work done when a force of one Newton causes a displacement of one meter.

🎯 Exam Tip: Understand that a Joule is derived from Newton-meter. Both terms (Newton-meter and Joule) correctly represent the SI unit of work.

Question 17.
......................... is the capacity to do work.
(a) Energy
(b) Force
(c) Power
(d) Momentum
Answer: (a) Energy
In simple words: Energy is the fundamental physical quantity that defines an object's or system's capacity to perform work or produce heat.

🎯 Exam Tip: The definition of energy as the capacity to do work is crucial. Don't confuse it with force (a push/pull) or power (rate of doing work).

Question 18.
Kinetic energy of a body (KE) = .........................
(a) mv²
(b) 1/2 mv²
(c) mgh
(d) Fs
Answer: (b) 1/2 mv²
In simple words: The formula for kinetic energy is one-half times the mass of the object multiplied by the square of its velocity, representing the energy it possesses due to its motion.

🎯 Exam Tip: Memorize the formula for kinetic energy, \( \frac{1}{2}mv^2 \). Pay attention to the square of velocity, as it means kinetic energy is very sensitive to speed changes.

Question 19.
Potential energy of a body is given by (P.E.) = .........................
(a) Fs
(b) mgh
(c) ma
(d) mv²
Answer: (b) mgh
In simple words: Gravitational potential energy (P.E.) is calculated by multiplying the mass (m) of an object by the acceleration due to gravity (g) and its height (h) above a reference point, representing stored energy due to position.

🎯 Exam Tip: Remember the formula for gravitational potential energy, \( mgh \). Understand that it depends on the object's mass, gravity, and height, not its motion.

Question 20.
1 hp = .........................
(a) 476 watts
(b) 746 watts
(c) 674 watts
(d) 764 watts
Answer: (b) 746 watts
In simple words: One horsepower (hp) is a common unit of power, especially for engines and motors, and it is equivalent to approximately 746 watts.

🎯 Exam Tip: Know common unit conversions, especially for power. The conversion factor between horsepower and watts is a frequently tested value.

Question 21.
......................... is the commercial unit of power.
(a) kilowatt second
(b) dyne
(c) kilowatt
(d) erg
Answer: (c) kilowatt
In simple words: Kilowatt (kW) is the commercial unit for power, often used to rate the output of large electrical appliances or generators. Kilowatt-hour (kWh) is the commercial unit for energy.

🎯 Exam Tip: Differentiate between commercial units for power (kilowatt) and energy (kilowatt-hour). This is important for practical applications.

Question 22.
1 kWh = ......................... joules.
(a) 3.6 x 10³
(b) 3.6 x 10⁶
(c) 6.3 x 10⁶
(d) 6.3 x 10³
Answer: (b) 3.6 x 10⁶
In simple words: One kilowatt-hour (kWh) is a unit of energy commonly used for billing electricity. It equals the energy consumed by a 1-kilowatt device operating for one hour, which converts to 3.6 million Joules.

🎯 Exam Tip: The conversion of 1 kWh to Joules (\( 3.6 \times 10^6 \text{ J} \)) is a standard value that should be memorized for calculations involving energy consumption.

Based On Practicals

Question 23.
The work done by a force is said to be ......................... when the applied force does not produce displacement.
(a) positive
(b) negative
(c) zero
(d) none of these
Answer: (c) zero
In simple words: If a force is applied to an object but the object does not move or undergo any displacement, then no work is done. This is a key condition for work to be zero.

🎯 Exam Tip: Reinforce the definition of work: force multiplied by displacement. If displacement is zero, work is always zero, regardless of the force applied.

Question 24.
When some unstable atoms break up, they release a tremendous amount of ......................... energy.
(a) chemical
(b) potential
(c) nuclear
(d) mechanical
Answer: (c) nuclear.
In simple words: The energy released when unstable atoms undergo radioactive decay or fission is called nuclear energy, resulting from changes within the atomic nucleus.

🎯 Exam Tip: Understand the different sources and forms of energy. Nuclear energy is unique in its origin from the atom's nucleus and its immense power output.

Name The Following:

Question 1.
Unit of energy used for commercial purpose.
Answer: Kilowatt-hour kW h is the unit of energy used for commercial purpose.
In simple words: The kilowatt-hour (kWh) is the practical unit used by electricity companies to measure and bill for the electrical energy consumed in homes and businesses.

🎯 Exam Tip: Distinguish between the SI unit of energy (Joule) and the commercial unit (kilowatt-hour) for practical applications.

Question 2.
Unit used in industry to measure power.
Answer: Horse power (hp) is the unit used in industry to express power.
In simple words: Horsepower (hp) is a traditional unit of power often used in industries to describe the output of engines, motors, and other machinery.

🎯 Exam Tip: Be aware of both SI units (Watt) and traditional units (horsepower) for power, especially their common uses in different contexts.

Question 3.
Sl unit of energy.
Answer: SI unit of energy is Joule (J).
In simple words: The Joule (J) is the standard international unit for all forms of energy, including kinetic, potential, and thermal energy.

🎯 Exam Tip: Always use Joule (J) as the primary unit for energy in scientific calculations and explanations unless specified otherwise.

Question 4.
Two types of mechanical energy.
Answer: Potential energy and kinetic energy are the two types of mechanical energy.
In simple words: Mechanical energy is the total energy an object possesses due to its motion (kinetic energy) and its position or configuration (potential energy).

🎯 Exam Tip: Mechanical energy is a combination of kinetic and potential energy. Understanding these two components is key to grasping mechanical systems.

Question 5.
An example where force acting on an object does not do any work.
Answer: In a simple pendulum, the gravitational force acting on the bob does not do any work as there is no displacement in the direction of force.
In simple words: When a force is applied perpendicular to the direction of an object's motion, it does no work on the object. For a pendulum, gravity acts vertically, but the pendulum's motion is horizontal at its lowest point, so gravity does no work then.

🎯 Exam Tip: Remember that work is only done when there is a component of force in the direction of displacement. If force and displacement are perpendicular, work is zero.

Question 6.
The relationship between 1 joule and 1 erg.
Answer: 1 joule = 10⁷ erg.
In simple words: A Joule is the SI unit of energy, while an erg is the CGS unit of energy. One Joule is equal to ten million ergs, indicating the Joule is a much larger unit.

🎯 Exam Tip: Know the conversion factor between SI and CGS units for work/energy (Joule and erg). This is a common conversion. \( 1 \text{ J} = 10^7 \text{ erg} \).

Question 7.
Various forms of energy
Answer: The various forms of energy are mechanical, heat, light, sound, electro-magnetic, chemical, nuclear and solar.
In simple words: Energy exists in many different forms, all capable of being converted into one another, such as energy of motion (mechanical), thermal (heat), radiant (light/solar), vibrational (sound), and stored energy (chemical/nuclear/electromagnetic).

🎯 Exam Tip: Be able to list and briefly describe different forms of energy. This shows a broad understanding of energy concepts.

State Whether The Following Statements Are True Or False:

Question 1.
(1) The potential energy of a body of mass 1 kg kept at height 1 m is 1 J.
Answer: (1) False
In simple words: The potential energy (mgh) of a 1 kg mass at 1 m height is \( 1 \times 9.8 \times 1 = 9.8 \text{ J} \), not 1 J, as 'g' is approximately 9.8 m/s².

🎯 Exam Tip: Always include the value of 'g' (acceleration due to gravity) when calculating potential energy. It's approximately 9.8 m/s² on Earth.

Question 2.
(2) Water stored at some height has potential energy.
Answer: (2) True
In simple words: Water held at a height, like in a dam, possesses gravitational potential energy because of its elevated position relative to the ground.

🎯 Exam Tip: Gravitational potential energy is stored energy due to an object's position in a gravitational field; water in a dam is a perfect example.

Question 3.
(3) Unit of power is joule.
Answer: (3) False
In simple words: The unit of power is Watt (W), which represents Joules per second (J/s), while Joule (J) is the unit for energy or work.

🎯 Exam Tip: Do not confuse units of energy (Joule) with units of power (Watt). Power is the rate at which energy is used or work is done.

Question 4.
(4) Mechanical energy can be converted into electrical energy.
Answer: (4) True
In simple words: Mechanical energy, such as that from rotating turbines in power plants, is commonly converted into electrical energy by generators.

🎯 Exam Tip: Energy conversion is a key concept. Generators are practical examples of converting mechanical energy into electrical energy.

Question 5.
(5) Work is a vector quantity.
Answer: (5) False
In simple words: Work is a scalar quantity because it only has magnitude, not direction, even though it is calculated from vector quantities like force and displacement.

🎯 Exam Tip: Differentiate between scalar (magnitude only) and vector (magnitude and direction) quantities. Work is scalar; force and displacement are vectors.

Question 6.
(6) Power is a scalar quantity.
Answer: (6) True
In simple words: Power, defined as the rate at which work is done or energy is transferred, only has magnitude and no directional component, making it a scalar quantity.

🎯 Exam Tip: Like work and energy, power is also a scalar quantity. It measures how much work is done per unit time, not in what direction.

Question 7.
(7) The kilowatt hour is the unit of energy.
Answer: (7) True
In simple words: Kilowatt-hour (kWh) is a unit of energy, often used for commercial electricity consumption, and it represents the energy equivalent of using one kilowatt of power for one hour.

🎯 Exam Tip: Understand that kilowatt-hour is a unit of energy, not power, despite having "kilowatt" in its name. It's a common point of confusion.

Question 8.
(8) The CGS unit of energy is dyne.
Answer: (8) False
In simple words: Dyne is the CGS unit of force, while the CGS unit of energy is erg.

🎯 Exam Tip: Be careful not to mix up units of force and energy in different systems. Dyne is for force, erg is for energy.

Question 9.
(9) The Sl unit of work is newton.
Answer: (9) False
In simple words: Newton (N) is the SI unit of force, whereas the SI unit of work is the Joule (J).

🎯 Exam Tip: Clearly distinguish between force (Newton) and work (Joule) and their respective SI units.

Question 10.
(10) Kinetic energy has formula – mv²
Answer: (10) True
In simple words: The formula \( \frac{1}{2}mv^2 \) represents kinetic energy, so \( mv^2 \) is proportional to it and directly related to the energy of motion.

🎯 Exam Tip: While \( \frac{1}{2}mv^2 \) is the full kinetic energy formula, stating \( mv^2 \) is still an incomplete formula for kinetic energy, but the statement "Kinetic energy has formula - \( mv^2 \)" can be considered true in a loose sense if referring to the dependency on mass and square of velocity. For strict accuracy, \( \frac{1}{2}mv^2 \) is correct.

Find The Odd Man Out.

Question 1.
Work, Energy, Power, Force.
Answer: Force.
In simple words: Work, Energy, and Power are all scalar quantities related to the transfer or rate of transfer of energy. Force is a vector quantity that causes a change in motion.

🎯 Exam Tip: Group concepts based on their fundamental nature (scalar/vector) or their relationship (e.g., work, energy, power are closely related). Force is distinct as it causes change in motion.

Question 2.
A stretched spring, A body placed in at some height, A bullet fired from gun.
Answer: A bullet fired from gun.
In simple words: A stretched spring and a body at height both represent potential energy (stored energy). A bullet fired from a gun represents kinetic energy (energy of motion).

🎯 Exam Tip: Identify the dominant type of energy present in each scenario (potential vs. kinetic). This helps categorize different physical states.

Question 3.
A stretched spring, A rock rolling downhill, A bullet fired from gun.
Answer: A stretched spring.
In simple words: A stretched spring stores potential energy due to its deformation. A rock rolling downhill and a bullet fired from a gun both primarily involve kinetic energy due to their motion.

🎯 Exam Tip: Carefully analyze the primary form of energy. A stretched spring stores potential energy, while rolling rocks and fired bullets are examples of kinetic energy.

Write The Formula Of The Following.

Question 1.
Kinetic energy
Answer: \( \frac{1}{2}mv^2 \)
In simple words: Kinetic energy is calculated by taking half of an object's mass multiplied by the square of its speed.

🎯 Exam Tip: Knowing fundamental formulas is essential. \( KE = \frac{1}{2}mv^2 \) relates kinetic energy to mass and velocity.

Question 2.
Potential energy
Answer: mgh
In simple words: Potential energy, specifically gravitational potential energy, is found by multiplying an object's mass by gravity and its height.

🎯 Exam Tip: \( PE = mgh \) is the formula for gravitational potential energy, dependent on mass, gravitational acceleration, and height.

Question 3.
Work
Answer: Fs or Fs cosθ
In simple words: Work is calculated as the product of force and displacement. If the force and displacement are not in the same direction, the cosine of the angle between them is also included in the formula.

🎯 Exam Tip: The work formula \( W = Fs \cos\theta \) is general. Use \( W = Fs \) when force and displacement are in the same direction (\( \cos 0^\circ = 1 \)).

Question 4.
Force
Answer: ma
In simple words: According to Newton's second law of motion, force (F) is the product of an object's mass (m) and its acceleration (a).

🎯 Exam Tip: Newton's second law, \( F = ma \), is a core principle of classical mechanics. Remember this fundamental relationship.

Question 5.
Power
Answer: \( \frac{W}{T} \)
In simple words: Power is defined as the rate at which work (W) is done, calculated by dividing the work done by the time (T) taken to do it.

🎯 Exam Tip: Power is a measure of how quickly work is performed. Its formula is \( P = \frac{W}{t} \), where W is work and t is time.

One Line Answer.

Question 1.
(i) When is work done said to be zero?
Answer: Work done is zero when force acting on the body and its displacement are perpendicular to each other.
(ii) Which quantities are measured in ergs?
Answer: Work and energy are measured in ergs.
(iii) What is the relationship between newton, meter and joule?
Answer: 1 joule = 1 newton x 1 meter
(iv) What is energy?
Answer: The ability of a body to do work is called energy.
(v) Give 4 examples of energy
Answer: Solar, wind, mechanical and heat.
(vi) Which device converts electrical energy into heat?
Answer: Electric water heater (Geyser) converts electrical energy into heat.
In simple words: This section summarizes key definitions and relationships: work is zero when force is perpendicular to displacement; ergs measure work and energy; a Joule is a Newton-meter; energy is the capacity to do work; examples include solar, wind, mechanical, and heat energy; and a geyser converts electrical to heat energy.

🎯 Exam Tip: These are quick recall questions that test foundational definitions and basic scientific facts. Memorize these direct answers for efficiency.

Question 2.
Find whether work is positive, negative or zero.
(a) Person moving along circle from A to B.
Answer: Work done is positive as direction of applied force and displacement are the same.
(b) Person completing one circle and returns to position A.
Answer: Work done is zero because there is no displacement for the person.
(c) Person pushing a car in the forward direction.
Ans, Work done is positive as the motion of car is in the direction of the applied force.
(d) A car coming downhill even after pushing it in the opposite uphill direction.
Ans, Work done is negative as the motion of car is in opposite direction of the applied force.
(e) Motion of the clock pendulum.
Answer: work done is zero as there is no displacement of the pendulum and it comes back to its original position.
In simple words: This set of examples demonstrates positive work when force aids motion (moving along a circle, pushing a car forward), negative work when force opposes motion (car coming downhill while pushed uphill), and zero work when there's no net displacement (completing a circle) or the force is perpendicular to instantaneous motion (pendulum).

🎯 Exam Tip: Practice identifying positive, negative, and zero work by considering the relative directions of the applied force and the resulting displacement in various scenarios.

Give Scientific Reasons:

Question 1.
A moving ball hits a stationary ball and displaces it.
Answer:
• The moving ball has certain energy.
• When it hits the stationary ball, the energy is transferred to the stationary ball, because of which it moves.
• Hence, a moving ball hits a stationary ball and displaces it.
In simple words: When a moving ball collides with a stationary one, its kinetic energy is transferred to the stationary ball, causing it to move and demonstrating the principle of energy transfer during collisions.

🎯 Exam Tip: This illustrates the concept of energy transfer and conservation of momentum in collisions. The moving object imparts its energy to the stationary one.

Question 5. An exploding firecracker lights up as well as makes a sound.
Answer:- The exploding firecracker converts the chemical energy stored in it into light and sound respectively. - Here, energy is converted from one type to another. - Hence, an exploding firecracker lights as well as makes a sound.In simple words: Firecrackers convert stored chemical energy into light and sound, demonstrating energy transformation from one form to others.

🎯 Exam Tip: When explaining energy transformations, clearly identify the initial and final forms of energy and the intermediate processes.

Question 6. Work done on an artificial satellite by gravity is zero while moving around the earth.
Answer:- When the artificial satellite moves around the earth in a circular orbit, gravitation force acts on it. - The gravitational force acting on the satellite and its displacement are perpendicular to each other. i.e. \(\theta\) = 90° - For \(\theta\) = 90°, work done is zero. [ W = Fs cos 90 = 0) - Hence, work done on an artificial satellite by gravity is zero while moving around the earth.In simple words: Gravity does no work on a satellite in a circular orbit because the gravitational force is always perpendicular to the satellite's displacement, meaning no energy is gained or lost by gravity.

🎯 Exam Tip: Remember that work done is zero when the force and displacement are perpendicular. This is a common concept for circular motion problems.

Difference Between :

Question 1. Work and Power:
Answer:

🎯 Exam Tip: Clearly differentiate between work (energy transfer) and power (rate of energy transfer), including their formulas and units, for full marks.

Question 2. Work and Energy:
Answer:

🎯 Exam Tip: Understanding that energy is a prerequisite for work, and work is the manifestation of energy transfer, is crucial.

Solve The Following:

Type - A

Formula:
W = Fs cos\(\theta\)
If force and displacement are in same direction, then \(\theta\) = 0°, and cos \(\theta\) = 1
If force and displacement are in opposite direction, then \(\theta\) = 180°, and cos \(\theta\) = -1
If force and displacement are perpendiculars, then \(\theta\) = 90°, and cos \(\theta\) = 0

Question 1. Pravin has applied a force of 100 N on an object, at an angle of 60° to the horizontal. The object gets displaced in the horizontal direction and 400 J work is done. What is the displacement of the object? (cos 60° = 1/2)
Answer:
To Find:
Displacement (s) = ?
Formula:
W = Fs cos \(\theta\)
Solution:
W = Fs cos \(\theta\)
400 = 100 \(\times\) s \(\times \frac{1}{2}\)

\(\implies \frac{400 \times 2}{100} = s\)

\(\implies 4 \times 2 = s\)

\(\implies s = 8\) m
The object will be displaced through 8 m.
In simple words: Given the work done, force, and angle, we use the work formula W = Fs cos\(\theta\) to calculate the displacement, finding it to be 8 meters.

🎯 Exam Tip: Always use the work formula W = Fs cos\(\theta\) and correctly substitute the angle \(\theta\) (in degrees or radians for cos function) when the force is not parallel to displacement.

Question 2. A force of 50 N acts on an object and displaces it by 2 m. If the force acts at an angle of 60° to the direction of its displacement, find the work done.
Answer:
Given:
Force (F) = 50 N
Displacement (s) = 2 m
Angle (\(\theta\)) = 60°
Formula:
W = Fs cos\(\theta\)
Solution:
W = 50 N \(\times\) 2 m \(\times\) cos 60°
W = 100 \(\times\) 0.5 (since cos 60° = 0.5)
W = 50 J
In simple words: When a 50 N force acts at 60 degrees to displace an object by 2 meters, the work done is calculated as 50 Joules using W = Fs cos\(\theta\).

🎯 Exam Tip: Ensure you know the common trigonometric values (like cos 0°, 30°, 45°, 60°, 90°, 180°) as they frequently appear in work-energy problems.

Question 3. Raj applied a force of 20 N and moved a book 40 cm in the direction of the force. How much was the work done by Raj?
Answer:
Given:
Force (F) = 20 N
Displacement (s) = 40 cm = 0.4 m
(Since force and displacement are in the same direction, \(\theta\) = 0°, cos 0° = 1)
Formula:
W = Fs cos\(\theta\)
Solution:
W = 20 N \(\times\) 0.4 m \(\times\) cos 0°
W = 20 \(\times\) 0.4 \(\times\) 1
W = 8 J
In simple words: Raj did 8 Joules of work by applying a 20 N force to move a book 40 cm in the same direction.

🎯 Exam Tip: Always convert all units to SI (e.g., cm to m) before performing calculations to avoid errors and ensure correct answer units.

Type -B

Formula:
1) W = K.E = \(1/2 mv^2\)
2) W = P.E = mgh
- W = P.E, W = K.E
1 km/hr = \(\frac{1000}{3600}\) m/s = \(\frac{5}{18}\) m/s

Question 4. A stone having a mass of 250 gm is falling from a height. How much kinetic energy does it have at the moment when its velocity is 2 m/s?
Answer:
Given:
Mass (m) = 250 g = \(\frac{250}{1000}\) kg = 0.25 kg
Velocity (v) = 2 m/s
To Find:
Kinetic energy (K.E) = ?
Formula:
K.E = \(\frac{1}{2} mv^2\)
Solution:
K.E = \(\frac{1}{2} mv^2\)
K.E = \(\frac{1}{2} \times 0.25 \times (2)^2\)
K.E = \(\frac{1}{2} \times 0.25 \times 4\)
K.E = 0.25 \(\times\) 2
K.E = 0.5 J
The kinetic energy of the stone is 0.5 J
In simple words: A 250g stone moving at 2 m/s has a kinetic energy of 0.5 Joules, calculated using the formula KE = \(1/2 mv^2\).

🎯 Exam Tip: Always ensure mass is in kilograms (kg) and velocity in meters per second (m/s) when calculating kinetic energy in Joules.

Question 5. 500 kg water is stored in the overhead tank of a 10 m high building. Calculate the amount of potential energy stored in the water.
Answer:
Given:
Mass (m) = 500 kg
Height (h) = 10 m
Acceleration due to gravity (g) = 9.8 m/s²
To Find:
Potential energy (P.E) = ?
Formula:
P.E. = mgh
Solution:
P.E = mgh
= 500 \(\times\) 9.8 \(\times\) 10
= 500 \(\times\) 98
= 49000 J
The P.E of the stored water is 49000 J
In simple words: The potential energy of 500 kg of water stored at a height of 10 m is 49000 Joules, calculated using the formula PE = mgh.

🎯 Exam Tip: For potential energy calculations, ensure mass is in kg, height in m, and use the standard value of g = 9.8 m/s² (unless specified otherwise).

Question 6. Calculate the work done to take an object of mass 20 kg to a height of 10 m. (g = 9.8 m/s²)
Answer:
Given:
Mass (m) = 20 kg
Acceleration due to gravity (g) = -9.8 m/s² (negative because displacement is against gravity)
Displacement (s) = (h) = 10 m.
To Find:
Work done (W) = ?
Formula:
(i) W = P.E = mgh
Solution:
W = mgh
= 20 \(\times\) (-9.8) \(\times\) 10
= -1960 J
The work done to take an object of mass 20 kg to a height of 10 m is -1960 J.
In simple words: To lift a 20 kg object 10 m high, the work done *by gravity* is -1960 Joules, indicating work is done against gravity.

🎯 Exam Tip: When an object is lifted against gravity, the work done *by gravity* is negative. The work done *by the lifting force* would be positive.

Question 7. A body of 0.5 kg thrown upwards reaches a maximum height of 5 m. Calculate the work done by the force of gravity during this vertical displacement.
Answer:
Given:
Mass (m) = 0.5 kg
Acceleration due to gravity (g) = -9.8 m/s² (negative as the body moves upwards, opposite to gravity)
Displacement (s) = 5 m.
To Find:
Work done (W) = ?
Formula:
W = P.E = mgh
Solution:
W = mgh
= 0.5 \(\times\) (-9.8) \(\times\) 5
= -24.5 J
The work done by the force of gravity is -24.5 joule.
In simple words: When a 0.5 kg body is thrown upwards 5m, gravity performs -24.5 Joules of work because its force opposes the upward displacement.

🎯 Exam Tip: The sign of work done by gravity depends on the direction of displacement relative to gravity. Upward motion implies negative work by gravity, while downward motion implies positive work.

Question 8. 1 kg mass has a kinetic energy of 2 joule. Calculate its velocity.
Answer:
Given:
Mass (m) = 1 kg
Kinetic Energy (K.E) = 2 J
To Find:
Velocity (v) = ?
Formula:
K.E = \(\frac{1}{2} mv^2\)
Solution:
K.E = \(\frac{1}{2} mv^2\)
2 = \(\frac{1}{2} \times 1 \times (v)^2\)

\(\implies 4 = (v)^2\)

\(\implies v^2 = 4\)

\(\implies v = \sqrt{4}\)

\(\implies v = 2\) m/s
The velocity is 2 m/s
In simple words: A 1 kg object with 2 Joules of kinetic energy is moving at 2 m/s, found by rearranging the KE = \(1/2 mv^2\) formula.

🎯 Exam Tip: When solving for velocity from kinetic energy, remember to take the square root of \(v^2\) and ensure units are consistent (J for KE, kg for m, m/s for v).

Question 9. A rocket of mass 100 tonnes is propelled with a vertical velocity 1 km/s. Calculate kinetic energy.
Answer:
Given:
Mass (m) = 100 tonnes = 100 \(\times\) 1000 kg = 100000 kg
Velocity (v) = 1 km/s = 1000 m/s
To Find:
Kinetic Energy (K.E) = ?
Formula:
K.E = \(\frac{1}{2} mv^2\)
Solution:
K.E = \(\frac{1}{2} mv^2\)
K.E = \(\frac{1}{2} \times 100000 \times (1000)^2\)
K.E = \(\frac{1}{2} \times 10^5 \times 10^6\)
K.E = 0.5 \(\times\) 10\(^{11}\) J
K.E = 5 \(\times\) 10\(^{10}\) J
The kinetic energy of the rocket is 5 \(\times\) 10\(^{10}\) J
In simple words: A rocket with a mass of 100 tonnes moving at 1 km/s has an enormous kinetic energy of 5 \(\times\) 10\(^{10}\) Joules.

🎯 Exam Tip: Pay close attention to unit conversions (tonnes to kg, km/s to m/s) in numerical problems, as they are a common source of error for large numbers.

Type - C

Formula:
1) Power = \(\frac{\text{work}}{\text{time}} = \frac{mgh}{t}\)
2) Electric power = \(\frac{\text{Electric energy consumed}}{\text{time}}\)
Power should be expressed in kW
Time should be expressed in hours
1 kWh = 1 unit

Question 10. Swaralee takes 20 s to carry a bag weighing 20 kg to a height of 5 m. How much power has she used?
Answer:
Given:
Mass (m) = 20 kg
Height (h) = 5 m
Time (t) = 20 s
Acceleration due to gravity (g) = 9.8 m/s²
To Find:
Power (P) = ?
Formula:
P = \(\frac{mgh}{t}\)
Solution:
P = \(\frac{20 \times 9.8 \times 5}{20}\)
P = 9.8 \(\times\) 5
P = 49 W
Power used by Swaralee is 49 W
In simple words: Swaralee used 49 Watts of power to lift a 20 kg bag to a height of 5 m in 20 seconds.

🎯 Exam Tip: For power calculations, ensure all units are in SI (kg, m, s, m/s²) to directly obtain power in Watts.

Write Notes On The Following:

Question 1. Derive the expression for potential energy.
Answer:
(i) To carry an object of mass 'm' to a height 'h' above the earth's surface, a force equal to 'mg' has to be used against the direction of the gravitational force.
(ii) The amount of work done can be calculated as follows:
Work = force \(\times\) displacement
\(\therefore\) W = mg \(\times\) h
\(\therefore\) W = mgh
(iii) The amount of potential energy stored in the object because of its displacement.
PE = mgh (W = P.E)
(iv) Displacement to height h causes energy equal to mgh to be stored in the object.
In simple words: Potential energy (PE) is the energy stored in an object due to its position or state. For an object lifted to a height 'h', the work done against gravity (mg) to raise it is stored as potential energy, given by PE = mgh.

🎯 Exam Tip: The derivation of potential energy highlights that the work done against a conservative force (like gravity) is stored as potential energy.

Question 2. When can you say that the work done is either positive, negative or zero?
Answer:- When the force and the displacement are in the same direction, the work done by the force is positive. - When the force and displacement are in the opposite directions, the work done by the force is negative. - When the applied force does not cause any displacement or when the force and the displacement are perpendicular to each other, the work done by the force is zero.In simple words: Work is positive when force aids motion, negative when it opposes motion, and zero when force causes no displacement or is perpendicular to motion.

🎯 Exam Tip: Understanding the relationship between the direction of force and displacement is key to determining if work is positive, negative, or zero, often summarized by W = Fs cos\(\theta\).

Question 3. Explain the relation between, the commercial and SI unit of energy.
Answer:
The commercial unit of energy is a kilowatt-hour (kWh) while the SI unit of energy is the joule. Their relation is
1 kWh = 1 kW \(\times\) 1 hr
= 1000 W \(\times\) 3600 s
= 3600000 J
(Watt \(\times\) Sec = Joule)
1 kWh = 3.6 \(\times\) 10\(^6\) J.
In simple words: The kilowatt-hour (kWh) is a larger, commercial unit of energy, equivalent to 3.6 million Joules, which is the standard SI unit for energy.

🎯 Exam Tip: Be able to convert between kilowatt-hour (commercial unit) and joule (SI unit) as this conversion is frequently tested.

Question 4. How is work calculated if the direction of force and the displacement are inclined to each other?
Answer:
If the direction of force and the displacement are inclined to each other then, we must convert the applied force into the force acting along the direction of displacement.
If \(\theta\) is angle between force and displacement, then force (\(F_1\)) in direction of displacement is
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दर्शाता है कि एक वस्तु पर बल (F) कैसे लगाया जाता है, और यह बल दो घटकों में विभाजित होता है: एक विस्थापन (s) की दिशा में (\(F_1\)) और दूसरा लंबवत। कोण \(\theta\) बल और विस्थापन के बीच का कोण है, और \(F_1\) बल का वह घटक है जो वस्तु के क्षैतिज विस्थापन में योगदान करता है, जिससे कार्य की गणना में cos \(\theta\) पद का उपयोग होता है।
\(\therefore\) W = \(F_1 \times s\)

\(\implies \text{cos } \theta = \frac{\text{base}}{\text{hypotenuse}}\)

\(\implies \text{cos } \theta = \frac{F_1}{F}\)

\(\implies F_1 = F \text{ cos } \theta\)
Substituting the value of \(F_1\) in equation 1
Thus, the work done by \(F_1\) is
W = F cos \(\theta\) s
\(\therefore\) W = Fs cos\(\theta\)
In simple words: When force and displacement are at an angle, work is calculated using W = Fs cos\(\theta\), where only the component of force acting in the direction of displacement (\(F \text{ cos }\theta\)) contributes to the work.

🎯 Exam Tip: The formula W = Fs cos\(\theta\) is fundamental for calculating work when the force and displacement are not perfectly aligned.

Complete The Flow Chart.

Question 1. Transformation of energy
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह फ्लोचार्ट ऊर्जा के विभिन्न रूपांतरणों को दर्शाता है। इसमें दिखाया गया है कि कैसे एक इंजन/पंखा यांत्रिक ऊर्जा को विद्युत ऊर्जा में परिवर्तित करता है, जो आगे प्रकाश, ऊष्मा या ध्वनि ऊर्जा में बदल सकती है। यह सौर सेल, प्राथमिक/द्वितीयक सेल, माइक्रोफोन और थर्मो-युग्म द्वारा ऊर्जा रूपांतरण को भी चित्रित करता है।
In simple words: The flowchart illustrates how different devices convert various forms of energy (mechanical, chemical, light, sound, heat) into other forms, primarily electrical energy, for different applications.

🎯 Exam Tip: When analyzing energy transformation flowcharts, identify the input and output energy forms at each stage and the device facilitating the conversion.

Question 2. Transformation of energy
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह फ्लोचार्ट सूर्य से शुरू होकर ऊर्जा के रूपांतरण को दर्शाता है। सौर ऊर्जा, प्रकाश संश्लेषण के माध्यम से रासायनिक ऊर्जा (कोयला) में बदल जाती है, जो ऊष्मा ऊर्जा (बॉयलर) पैदा करती है। यह ऊष्मा टरबाइन को चलाकर यांत्रिक ऊर्जा उत्पन्न करती है, जो जनरेटर के माध्यम से विद्युत ऊर्जा में परिवर्तित होती है, और अंततः धारा में गतिज ऊर्जा के रूप में सामने आती है।
In simple words: This flowchart details the energy conversion chain from solar energy (photosynthesis) to chemical energy (coal), then to heat, mechanical, electrical, and finally kinetic energy (stream).

🎯 Exam Tip: Flowcharts on energy transformations require understanding the sequence of energy changes and the role of each component in the system.

Write Effects Of The Following With Examples.

Question 1. Force
Answer:- A force can move a stationary object. The force of engine makes a stationery car to move. - A force can stop a moving object. The force of brakes can stop a moving car. - A force can change the speed of a moving object. When a hockey player hits a moving ball, the speed of ball increases. - A force can change the direction of a moving object. In the game of carrom, when we take a rebound then the direction of striker changes because the edge of the carrom board exerts a force on the strike. - A force can change the shape and size of an object. The shape of kneaded wet clay changes when a potter converts it into pots of different shapes and sizes because the potter applies force on the kneaded wet clay.In simple words: Force can initiate motion, stop moving objects, alter their speed or direction, and even change an object's shape or size.

🎯 Exam Tip: When describing the effects of force, provide a clear example for each effect to illustrate your understanding.

Give Two Examples In Each Of The Following Cases:

Question 1. Potential energy
Answer:- Water stored in a dam - A compressed springIn simple words: Potential energy is stored energy; examples include water held behind a dam or a coiled spring.

🎯 Exam Tip: For potential energy, focus on examples where energy is stored due to position (like height) or state (like compression/stretching).

Question 2. Kinetic energy
Answer:- Water flowing - Bullet fired from a gunIn simple words: Kinetic energy is the energy of motion; examples include flowing water or a fired bullet.

🎯 Exam Tip: For kinetic energy, provide examples that clearly show objects in motion and possessing energy due to that movement.

Question 3. Chemical energy
Answer:- Chemical in cell - Explosive mixture of a bombIn simple words: Chemical energy is stored in molecular bonds; examples are a battery cell or an explosive.

🎯 Exam Tip: Chemical energy examples should involve substances that release energy through chemical reactions.

Question 4. Zero work done
Answer:- A stone tied to a string and whirled in a circular path - Motion of the earth and other planets moving around the sunIn simple words: Work done is zero when the force is perpendicular to displacement, as seen in uniform circular motion like a whirling stone or planetary orbits.

🎯 Exam Tip: Zero work done occurs either when there's no displacement or when the force and displacement are at a 90-degree angle to each other.

Question 5. Negative work done
Answer:- A cyclist applies brakes to his bicycle, but the bicycle still covers some distance. - When a body is made to slide on a rough surface, the work done by the frictional force.In simple words: Negative work occurs when the force opposes the direction of motion, like braking a bicycle or friction acting on a sliding object.

🎯 Exam Tip: Negative work implies that the force is acting to slow down or resist the motion of an object, often removing energy from the system.

Question 6. Positive work done
Answer:
(i) A boy moving from the ground floor to the first floor.
(ii) A fruit falling down from the tree.
In simple words: Positive work occurs when the force acts in the same direction as the displacement, such as lifting an object or a fruit falling due to gravity.

🎯 Exam Tip: Positive work indicates that the force is contributing to the motion or increasing the energy of the object.

Question 7. A 25 W electric bulb is used for 10 hours every day. How much electricity does it consume each day?
Answer:
Given:
Power (P) = 25 W = 25/1000 kW = 0.025 kW
Time (t) = 10 hrs
To Find:
Electric energy consumed = ?
Formula:
Electric energy consumed = power \(\times\) time
Solutions:
Electric energy consumed = power \(\times\) time
= 0.025 \(\times\) 10
= 0.25 kWh
The electric bulb consumes 0.25 kWh of electricity each day.
In simple words: A 25W bulb used for 10 hours daily consumes 0.25 kilowatt-hours (kWh) of electricity per day.

🎯 Exam Tip: When calculating electricity consumption, always convert power to kilowatts (kW) and time to hours to get the answer in kilowatt-hours (kWh), the standard commercial unit.

Question 8. If a TV of rating 100W is operated for 6 hrs per day, find the amount of energy consumed in any leap year?

Answer:
Given:
Power (P) = 100 W
= \( \frac{100}{1000} \) kW
= 0.1 kW
Time (t) = 6 × 366
= 2196 hrs.
To Find:
Electric energy consumed
Formula:
Electric energy consumed = power x time
Solution:
Electric energy consumed = power x time
= 0.1 x 2196
= 219.6 kWh
The amount of energy consumed is 219.6 kWh
In simple words: To calculate the energy consumed, first convert the power to kilowatts, then multiply it by the total operating time in hours for a leap year (366 days x 6 hours/day).

🎯 Exam Tip: Remember to convert power to kilowatts and time to hours for energy consumption calculations, especially for problems involving commercial units like kWh. Also, pay attention to the year (leap or common) for accurate total time. Use correct units for the final answer.

Complete The Paragraph

Question 1. ................ is the measure of energy transfer when a force (F) moves an object through a ................ (d). So when ................ is done, energy has been transferred from one energy store to another, and so: energy transferred = ................ done. Energy transferred and work done are both measured in ................ (J)

Answer: Work is the measure of energy transfer when a force (F) moves an object through a distance (d). So when work is done, energy has been transferred from one energy store to another, and so: energy transferred = work done. Energy transferred and work done are both measured in joules (J).
In simple words: Work measures energy transfer when a force causes displacement. This transferred energy, also known as work done, is measured in joules.

🎯 Exam Tip: For fill-in-the-blanks, understand the definition of work and energy and their relationship. Know the SI unit of energy and work (Joule) and how it relates to force and distance.

Question 2. ................ energy and ................ done are the same thing as much as ................ energy and work done are the same thing. Potential energy is a state of the system, a way of ................ energy as of virtue of its configuration or motion, while ................ done in most cases is a way of channeling this energy from one body to another.

Answer: Potential energy and work done are the same thing as much as kinetic energy and work done are the same thing. Potential energy is a state of the system, a way of storing energy as of virtue of its configuration or motion, while work done in most cases is a way of channeling this energy from one body to another.
In simple words: Potential energy, kinetic energy, and work done are all forms of energy. Potential energy is stored due to position or state, while kinetic energy is due to motion. Work done is how energy is transferred between objects.

🎯 Exam Tip: Distinguish clearly between potential energy (stored energy), kinetic energy (energy of motion), and work (energy transfer). Emphasize that these are different manifestations or aspects of energy.

Question 3. In physics, ................ is the rate of doing work or, i.e., the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the ................ equal to one ................ per second.
Power is a ................ quantity that requires both a change in the physical system and a specified time interval in which the change occurs. But more ................ is needed when the work is done in a shorter amount of time.

Answer: In physics, power is the rate of doing work or, i.e., the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt. equal to one joule per second.
In simple words: Power is the speed at which work is done or energy is transferred. The SI unit for power is the watt, which means one joule per second.

🎯 Exam Tip: Know the definition of power as the rate of doing work. Be familiar with the SI unit of power (Watt) and its equivalence to joule per second. Understand that higher power means work is done faster.

Activity-based Questions

Answer In Detail:

Question 1. State the expression for work done when displacement and force makes an angle \( \theta \) OR State the expression for work done when force is applied making an angle \( \theta \) with the horizontal force.

Answer: Let 'F' be the applied force and \( F_{1} \) be its component in the direction of displacement. Let 'S' be the displacement.
The amount of work done is given by W = \( F_{1}s \) ............ (1)
The force 'F' is applied in the direction of the string.
Let ' \( \theta \)' be the angle that the string makes with the horizontal. We can determine the component ' \( F_{1} \)', of this force F, which acts in the horizontal direction by means of trigonometry.
\( F_{1} \) = F cos \( \theta \)
Substituting the value of \( F_{1} \) in equation 1
Thus, the work done by \( F_{1} \) is
W cos \( \theta \) s
\( \implies \) W = Fscos \( \theta \)
In simple words: When a force is applied at an angle to the displacement, only the component of the force parallel to the displacement contributes to the work done. This is expressed as W = Fs cos \( \theta \), where \( \theta \) is the angle between the force and displacement.

🎯 Exam Tip: Understanding the component of force is crucial for work-done problems involving angles. Always use W = Fs cos \( \theta \) and correctly identify the angle between the force and displacement vectors. Remember that only the force component parallel to displacement does work.

Question 2. When a body is dropped on the ground from some height its P.E is converted into K.E but when it strikes the ground and it stops, what happens to the K.E?

Answer: When a body is dropped on the ground, its K.E appears in the form of:
(i) Heat (collision between the body and the ground).
(ii) Sound (collision of the body with the ground).
(iii) The potential energy of change in state of the body and the ground.
(iv) Kinetic energy is also utilized to do work i.e., the ball bounces to a certain height and moves to a certain distance vertically and horizontally till Kinetic energy becomes zero.
(v) The process in which the kinetic energy of a freely falling body is lost in an unproductive chain of energy is called the dissipation of energy.
In simple words: When a falling object hits the ground and stops, its kinetic energy is not destroyed. Instead, it transforms into other forms of energy such as heat (due to friction and deformation), sound, and can cause work to be done (like bouncing or deforming the ground), a process known as energy dissipation.

🎯 Exam Tip: This question tests the understanding of the Law of Conservation of Energy, specifically energy transformation during impact. List all possible forms of energy conversion (heat, sound, work, deformation) and introduce the concept of energy dissipation.

Question 3. Explain the statement "Potential Energy is relative”.

Answer:
(i) The potential energy of an object is determined and calculated according to a height of the object with respect to the observer.
(ii) So, the person staying on 6th floor more potential energy than those staying on the 3rd floor.
(iii) But, the person on the 6th floor will have lesser potential energy than on the 8th floor. Hence potential energy is relative.
In simple words: Potential energy is relative because its value depends on the chosen reference point or observer. An object's potential energy can be different when measured from different heights.

🎯 Exam Tip: Emphasize that potential energy is not an absolute value but depends on the chosen zero reference level. Provide examples that illustrate how changing the reference point changes the calculated potential energy.