GSEB Class 9 Science Solutions Chapter 11 Work and Energy

Question 1. A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (fig. 11.1.). Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Answer: Given, the applied force is \( F = 7 \, N \). The displacement is \( s = 8 \, m \). Work done \( W \) is calculated as \( W = F \times s \). Therefore, \( W = 7 \times 8 = 56 \, J \). The work performed in this situation is \( 56 \, J \).
In simple words: A force of 7 Newtons pushes an object 8 meters. To find the work done, you multiply the force by the distance, which gives 56 Joules.

Exam Tip: Remember that work is done only when there is a displacement in the direction of the applied force.

Question 1. When do we say that work is done?
Answer: Work is considered to be done when specific conditions are fulfilled. First, a force must act on an object. Second, the object needs to experience a displacement. If these two requirements are met, then work is performed.
In simple words: We say work is done if a force pushes or pulls something, and that something actually moves because of it.

Exam Tip: To score full marks, clearly state both conditions required for work to be considered done in physics.

Question 2. Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer: The expression for work done when a force acts in the direction of an object's displacement is \( W = Fs \). Here, \( F \) stands for force, and \( s \) represents displacement.
In simple words: The formula for work is Force times displacement (W=Fs). 'F' is the pushing or pulling, and 's' is how far it moves.

Exam Tip: Memorize the basic work formula \( W = Fs \) and be able to define each variable clearly.

Question 3. Define 1 J of work.
Answer: We understand that work done \( W = Fs \). If a force of \( F = 1 \, N \) moves an object by \( s = 1 \, m \), then \( 1 \, J = 1 \, Nm \). Therefore, one joule (1 J) of work happens when a 1 Newton (N) force causes an object to move 1 meter (m) in the direction of that force.
In simple words: One Joule of work is done when you push something with 1 Newton of force and it moves 1 meter in that direction.

Exam Tip: Be precise in defining units. Ensure you mention both the force (Newtons) and the distance (meters) for a Joule.

Question 4. A pair of bullocks exerts a force of 140 N on a plow. The field being plowed is 15 m long. How much work is done in plowing the length of the field?
Answer: Given: The force applied by the bullocks is \( F = 140 \, N \). The displacement of the plow is \( s = 15 \, m \). To calculate work done, \( W = F \times s \). So, \( W = 140 \times 15 = 2100 \, J \). The work performed in plowing the field is \( 2100 \, J \).
In simple words: The bullocks use 140 Newtons of force to plow a 15-meter field. To find the total work done, multiply the force by the distance, which gives 2100 Joules.

Exam Tip: Always write down the given values and the formula before calculating to show your understanding and gain partial marks.

Question 1. What is power?
Answer: Power is defined as the speed at which work is done or the speed at which energy is transferred. It can be calculated as Power \( = \) work / time.
In simple words: Power tells us how fast work is done or how quickly energy moves from one place to another.

Exam Tip: Clearly state that power is a *rate* – either the rate of doing work or the rate of energy transfer.

Question 2. Write an expression for the kinetic energy of an object.
Answer: The expression for an object's kinetic energy is \( EK = \frac{1}{2}mu^2 \). Here, \( m \) represents the mass of the object, and \( u \) denotes its velocity.
In simple words: The formula for kinetic energy is one-half times mass times velocity squared. 'm' is mass, and 'u' is speed.

Exam Tip: Ensure you correctly write the formula and identify all variables with their proper units.

Question 3. The kinetic energy of an object of mass, m moving with a velocity of 5 m/s is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Answer: Initially, the kinetic energy (KE) of an object with mass \( m \) moving at \( 5 \, m/s \) is \( 25 \, J \). We know \( KE = \frac{1}{2}mv^2 \).
So, \( 25 = \frac{1}{2}m(5)^2 \)
\( 25 = \frac{1}{2}m(25) \)
\( m = \frac{2 \times 25}{25} = 2 \, kg \). When the velocity is doubled, it becomes \( v = 10 \, m/s \).
The new kinetic energy is \( KE = \frac{1}{2}m(10)^2 = \frac{1}{2}(2)(100) = 100 \, J \). When the velocity is increased three times, it becomes \( v = 15 \, m/s \).
The kinetic energy will be \( KE = \frac{1}{2}m(15)^2 = \frac{1}{2}(2)(225) = 225 \, J \).
In simple words: First, we found the object's mass using its initial energy and speed. Then, we used that mass to calculate the new kinetic energy when the speed was doubled, and again when it was tripled.

Exam Tip: Remember that kinetic energy is proportional to the square of velocity, so doubling velocity quadruples KE, and tripling velocity increases KE by nine times.

Question 1. What is the kinetic energy of an object?
Answer: Kinetic energy refers to the energy that a body possesses because of its movement. It is the energy an object has when it is moving.
In simple words: Kinetic energy is the energy an object has because it is moving.

Exam Tip: Emphasize that kinetic energy is directly linked to motion and is not present in stationary objects.

Question 2. Define 1 Watt of power.
Answer: We know that power \( P = w/t \), where \( w \) is work and \( t \) is time. If \( w = 1 \, J \) and \( t = 1 \, s \), then \( P = \frac{1J}{1s} \). Thus, \( 1 \, W = 1 \, J/s \). One watt (1 W) is the power of an agent that performs work at a rate of 1 joule per second.
In simple words: One Watt is when you do 1 Joule of work in 1 second.

Exam Tip: State the formula \( P = W/t \) and then substitute the values for one Joule and one second to derive the definition of one Watt.

Question 3. A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Answer: Given: The electrical energy consumed is \( E = 1000 \, J \). The time taken is \( t = 10 \, s \). Power \( P \) can be calculated as \( P = \frac{W}{t} \) or \( P = \frac{E}{t} \). So, \( P = \frac{1000 \, J}{10 \, s} \). Therefore, the power of the lamp is \( P = 100 \, W \).
In simple words: A lamp uses 1000 Joules of energy in 10 seconds. To find its power, divide the energy by the time, which gives 100 Watts.

Exam Tip: Remember that power can be calculated using either work or energy divided by time, as work done is equivalent to energy transferred.

Question 4. Define average power.
Answer: Average power is the proportion of the total work completed to the total time spent, or the proportion of the total energy used to the total time taken. This means \( P_{avg} = \frac{\text{Total work done}}{\text{Total time}} = \frac{\text{Total energy consumed}}{\text{Total time}} \).
In simple words: Average power is the total work done divided by the total time it took, or the total energy used divided by the total time.

Exam Tip: Emphasize "total" for both work/energy and time when defining average power, differentiating it from instantaneous power.

In-Text Activities Solved

Activity 11.4 Discussion and Conclusion

• Work done by an external force is positive as the displacement happens in the same direction as the force.
• Work done by the gravitational force is negative when gravity acts downwards and displacement occurs upwards.

Activity 11.5 Discussion and Conclusion

• Types of energy include fossil fuels, solar power, nuclear power, wind power, hydro power, ocean power, biomass, and others.
• Hydro-energy is the power from moving water, which comes from the natural water cycle driven by solar energy. Wind energy comes from air movement in the atmosphere, caused by the uneven heating of land and sea by solar radiation.

Activity 11.6 Discussion and Conclusion

1. When a ball drops, its potential energy changes into kinetic energy. Eventually, the resistive force from the sand makes the ball stop.

2. The ball ceases movement because of the sand's frictional force.

Therefore, \( mgh = F \times x \)
Where \( F \) represents the friction force present.
\( x \) is the depth of the mark made in the sand bed.
\( m \) stands for the ball's mass.
\( h \) is the height from where the ball is released.
This implies \( x \propto h \).
If the ball is released from a greater elevation, the spread will be larger. Consequently, the depression will be deepest when the object is dropped from \( 1.5 \, m \) and shallowest when dropped from \( 25 \, cm \).

For height \( h = 4 \, m \),
\( E_p = mgh = 20 \times 10 \times 4 = 800 \, J \)
\( E_k = 0 \)

For height \( h = 3 \, m \),
\( E_p = mgh = 20 \times 10 \times 3 = 600 \, J \)
\( E_k = \frac{1}{2}mu^2 = mgx = 20 \times 10 \times 1 = 200 \, J \)

For height \( h = 2 \, m \),
\( E_p = mgh = 20 \times 10 \times 2 = 400 \, J \)
\( E_k = \frac{1}{2}mu^2 = mgx = 20 \times 10 \times 2 = 400 \, J \)

For height \( h = 1 \, m \),
\( E_p = mgh = 20 \times 10 \times 1 = 200 \, J \)
\( E_k = \frac{1}{2}mu^2 = mgx = 20 \times 10 \times 3 = 600 \, J \)

For height \( h = \) Just above the ground
\( E_p = mgh = 20 \times 10 \times 0 = 0 \, J \)
\( E_k = \frac{1}{2}mu^2 = mgx = 20 \times 10 \times 4 = 800 \, J \)

The work performed is calculated as Force multiplied by displacement. So, \( W = F \times h = mgh = 80 \, mJ \).

Child A performed work per unit of time. \( P_A = \frac{W_A}{t_A} = \frac{80 \, mJ}{15 \, s} \).

Child B performed work per unit of time. \( P_B = \frac{W_B}{t_B} = \frac{80 \, mJ}{20 \, s} \).

Here, \( P_A > P_B \), which means Child A's power output is greater than Child B's power output.

The activity demonstrates two key concepts: first, the conservation of mechanical energy in a falling object, where the total sum of potential and kinetic energy remains constant. Second, it shows that power is the rate at which work is done; if two individuals do the same amount of work but one does it faster, that individual has more power.

Activity 11.7 Discussion and Conclusion

1. In this scenario, the potential energy from the mass in the pan changes into kinetic energy for the trolley. This energy is then ultimately passed on to the block.

2. If the mass in the pan gets bigger, more potential energy converts into kinetic energy. As a result, the block's displacement will be greater.

Activity 11.8 Discussion and Conclusion

While stretching the rubber band, we perform work against the restoring force. This work becomes stored potential energy (elastic potential energy). When we let go of one end, this elastic potential energy changes into kinetic energy.

Activity 11.9 Discussion and Conclusion

• A Slinky naturally tries to return to its initial shape.
• The energy used in stretching a Slinky is stored as spring potential energy (elastic potential energy).
• Yes, the Slinky would also gain elastic potential energy when it is squashed.

Activity 11.10 Discussion and Conclusion

• Yes, it shifts when placed on the surface.
• It gains power from the spring's elastic potential energy.
• Indeed, the toy's acquired energy depends on how many times the spring is wound. Winding the toy's spring more makes it store more potential energy.
• To verify this, wind the toy's spring five times and measure the distance it travels on the ground. Then, wind the spring ten times. The toy will then cover a significantly greater distance.

Activity 11.11 Discussion and Conclusion

When we work against the gravitational field to lift a mass, this work done becomes gravitational potential energy.

Activity 11.12 Discussion and Conclusion

When the arrow is released from the bow, the elastic potential energy stored in the bow and stretched string changes into the kinetic energy of the arrow, causing the bow to regain its original shape.

Activity 11.13 Energy Conversions

1. Electrical energy changes into sound energy: Examples include an electric bell, a stereo system, or a loudspeaker.

2. Heat energy is saved by changing into mechanical energy, as seen in heat engines.

(a) Green plants make food through the process of photosynthesis.

(b) Plants receive energy directly from the Sun.

(c) Air travels from one area to another because of differences in pressure. These pressure changes occur due to the uneven heating of the land and sea by solar radiation.

(d) Fossil fuels developed millions of years ago, as plant and animal remains were buried deep underground and exposed to high temperatures and pressures. Solar energy changes water into water vapor, which then increases the water's thermal energy. Because of the buoyant force on water vapor, its potential energy grows. This water vapor then forms clouds. When clouds release rain, their potential energy transforms into the kinetic energy of the falling raindrops.

Gujarat Board Class 9 Science Work and Energy Textbook Questions and Answers

Question 1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term 'work'.
(a) Suma is swimming in a pond.
(b) A donkey is carrying a load on its back.
(c) A windmill is lifting water from a well.
(d) A green plant is carrying out photosynthesis.
(e) An engine is pulling a train.
(f) Food grains are getting dried in the Sun.
(g) A sailboat is moving due to wind energy.
Answer:
(a) Suma is performing work as she propels herself by using muscle force with her arms and legs in the water.
(b) A donkey is not doing any work in this situation because the force (weight) and its displacement are perpendicular to each other.
(c) A windmill does work against gravity to raise water. Therefore, work is being done by the windmill as it lifts water from the well.
(d) In this instance, the plant's leaves do not move. As a result, the work done is zero.
(e) An engine exerts force to pull the train, causing it to move in the same direction as the force. This results in a displacement of the train in the force's direction, meaning the engine performs work on the train.
(f) When food grains dry in the sun, solar radiation removes moisture. Since there is no force or displacement is involved in this process, the work done is zero.
(g) Wind energy exerts a force on the sailboat, moving it forward. This creates a displacement of the boat in the direction of the force, indicating that the wind performs work on the boat.
In simple words: Work is done when a force causes movement in the same direction. For Suma, windmill, engine, and sailboat, work happens. For the donkey, plant, and drying grains, no work is done because there's no movement in the force's direction.

Exam Tip: For each scenario, explicitly state whether work is done and provide a brief justification based on the definition of work (force and displacement in the same direction).

Question 2. An object was thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer: The work done by gravity on an object depends solely on its vertical displacement. In this case, the vertical displacement is the difference between the object's initial and final positions, which is zero. Therefore, no work is done. The horizontal movement is perpendicular to the direction of gravity.
In simple words: When an object lands at the same height it started from, gravity does no work because there's no overall change in its vertical position.

Exam Tip: Always remember that gravitational work depends only on the vertical change in height, not the path taken. If the vertical displacement is zero, the work done by gravity is zero.

Question 3. A battery lights a bulb. Describe the energy changes involved in the process.
Answer: A battery transforms chemical energy into electrical energy. This electrical energy is then changed into light energy when the bulb illuminates, along with some heat energy. The energy transformation sequence is: Chemical energy \( \implies \) Electrical energy \( \implies \) Light energy \( + \) Heat energy.
In simple words: A battery's chemical energy turns into electricity. This electricity then makes the bulb glow, producing both light and heat.

Exam Tip: When describing energy transformations, list the initial energy form, all intermediate forms, and the final energy forms, including any unintended forms like heat.

Question 4. The certain force acting on a 20 kg mass changes its velocity from 5 ms\(^{-1}\) to 2 ms\(^{-1}\). Calculate the work done by the force.
Answer: Given: The mass of the body is \( m = 20 \, kg \). The initial velocity is \( u = 5 \, m/s \). The final velocity is \( v = 2 \, m/s \).
The initial kinetic energy is \( KE_i = \frac{1}{2}mu^2 = \frac{1}{2} \times 20 \times (5)^2 \)
\( KE_i = 10 \times 25 = 250 \, J \).
The final kinetic energy is \( KE_f = \frac{1}{2}mv^2 = \frac{1}{2} \times 20 \times (2)^2 = 10 \times 4 = 40 \, J \).
Work done is equal to the change in kinetic energy, so \( W = KE_f - KE_i = 40 - 250 = -210 \, J \). The work done is \( -210 \, J \). The negative sign indicates a retarding force is acting.
In simple words: First, we calculated the object's kinetic energy before and after the force acted. Then, we subtracted the initial energy from the final energy to find the work done, which was negative because the force slowed it down.

Exam Tip: Remember the work-energy theorem: Work done is equal to the change in kinetic energy. A negative work done implies the force opposed the motion.

Question 5. mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Answer: The gravitational force performs zero work on the object. This occurs because the force of gravity acts downwards, while the object's displacement is horizontal, making them perpendicular to each other. As work is calculated by force times displacement in the direction of the force, no work is done when force and displacement are perpendicular.
In simple words: Gravity does no work when an object moves straight across because gravity pulls down, but the object moves sideways, making them perpendicular.

Exam Tip: Clearly state that work done by a force is zero if the force is perpendicular to the displacement. Use this principle for gravitational force on horizontally moving objects.

Question 6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer: No, this process does not go against the law of conservation of energy. As an object falls from a height, its potential energy gradually transforms into kinetic energy. The reduction in potential energy directly matches the gain in kinetic energy. Throughout this event, the body's total mechanical energy stays the same. Thus, the law of conservation of energy is not broken.
In simple words: No, it doesn't break the law of conservation of energy. As an object falls, its stored energy turns into motion energy, keeping the total energy the same.

Exam Tip: Explain that the decrease in potential energy is balanced by an equal increase in kinetic energy, ensuring that the total mechanical energy remains constant, thus upholding the law of conservation.

Question 7. What are the various energy transformations that occur when you are riding a bicycle?
Answer: While riding a bicycle, the muscular energy of the rider transforms into heat energy (which warms the rider's body) and also into the kinetic energy of the bicycle, providing motion. The transformation can be shown as: Muscular energy \( \implies \) Kinetic energy \( + \) Heat energy. Throughout this transformation, the total energy remains conserved.
In simple words: When you ride a bike, your muscle energy turns into the bike's movement energy and also heat energy, but the total energy stays the same.

Exam Tip: List all forms of energy involved, including mechanical (kinetic) and thermal (heat) energy, and mention the conservation of total energy.

Question 8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer: When we attempt to move a large rock without success, the energy we expend gets absorbed by the rock's particles. This causes an increase in the potential energy of the rock's particles and leads to deformation. However, this change is not noticeable because of the rock's enormous size.
In simple words: Yes, energy is transferred. When you push a huge rock and it doesn't move, your energy makes the rock's particles gain potential energy and deform slightly, though you can't see it.

Exam Tip: Explain that even without visible movement, internal changes like increased potential energy and microscopic deformation occur when a significant force is applied.

Question 9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Answer: One unit of energy is equivalent to 1 kilowatt-hour (kWh). Since \( 1 \, kWh = 3.6 \times 10^6 \, J \), then \( 250 \) units of energy consumed equals \( 250 \times 3.6 \times 10^6 = 9 \times 10^8 \, J \). So, the household consumed \( 9 \times 10^8 \, J \) of energy.
In simple words: A household used 250 energy units. Since one unit is 3.6 million Joules, the total energy used is 900 million Joules.

Exam Tip: Clearly state the conversion factor from kWh to Joules and perform the multiplication correctly to arrive at the final answer in Joules.

Question 10. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.
Answer: Given: The object's mass is \( m = 40 \, kg \). The vertical displacement (height) is \( h = 5 \, m \).
The potential energy (PE) of the mass is \( PE = mgh \)
\( PE = 40 \times 9.8 \times 5 = 1960 \, J \).
During free fall, its potential energy transforms into kinetic energy. At half the original height, the potential energy remaining is \( PE_{half} = mgh_{half} = 40 \times 9.8 \times 2.5 = 980 \, J \).
By the law of conservation of energy, the kinetic energy at half-height will be \( KE_{half} = PE_{initial} - PE_{at \, half-height} = 1960 \, J - 980 \, J = 980 \, J \).
In simple words: We first found the total stored energy at 5 meters. Then, at half the height, half of that stored energy would have turned into motion energy, which is 980 Joules.

Exam Tip: Use the law of conservation of mechanical energy, where the sum of potential and kinetic energy remains constant in the absence of non-conservative forces like air resistance.

Question 11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Answer: When a satellite orbits the Earth, the gravitational force acts towards the center of the Earth, and its direction is always perpendicular to the satellite's displacement at any moment. Because the force and displacement are at right angles, the work done by gravitational force is zero.
In simple words: Gravity does no work on a satellite circling Earth because gravity pulls it towards Earth's center, while the satellite moves sideways to that pull, making them perpendicular.

Exam Tip: To justify, explicitly state that work is zero because the force (gravity) and displacement are perpendicular to each other in circular orbit.

Question 12. Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
Answer: Yes, it is possible for an object to experience displacement even when no net force is acting upon it. This occurs when a body moves at a constant velocity. In such a scenario, although there is no net force, the object still undergoes displacement due to its continuous motion.
In simple words: Yes, an object can move without a force if it's already moving at a steady speed. It will keep moving because no force is stopping it.

Exam Tip: Refer to Newton's First Law of Motion, which states that an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

Question 13. A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Answer: When a person holds a bundle of hay above their head, no displacement occurs for the bundle. Therefore, the work done on the bundle of hay is zero, according to the physics definition of work. However, the person still experiences fatigue due to muscular effort required to hold the bundle against gravity.
In simple words: No work is done on the hay because it doesn't move. The person gets tired because their muscles are working to hold it up, but no physical work is done on the hay itself.

Exam Tip: Distinguish between physical work (force causing displacement) and biological effort (muscular activity). For work to be done, there must be displacement in the direction of the force.

Question 14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Answer: Given: The power rating of the electric heater is \( P = 1500 \, W \). The time for which the heater runs is \( t = 10 \, h \). The energy consumed by the electric heater is calculated using the formula \( E = P \times t \).
Therefore, \( E = 1500 \, W \times 10 \, h = 15000 \, Wh \). Converting this to kilowatt-hours, \( 15000 \, Wh = 15 \, kWh \). So, the heater uses \( 15 \) units of energy.
In simple words: The heater uses 1500 Watts of power for 10 hours. To find the total energy, multiply power by time, giving 15,000 Watt-hours or 15 kilowatt-hours.

Exam Tip: Ensure units are consistent; convert Watts to kilowatts or hours to seconds if necessary for specific energy units like Joules. Here, kWh is typically used for household energy consumption.

Question 15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Answer: When the pendulum bob is pulled, the energy given to it is kept as potential energy because of its higher location. When the pendulum is let go so it begins to move to the right, its potential energy changes into kinetic energy. At the middle point, it has the most kinetic energy and no potential energy.
As the pendulum moves to the far right, its kinetic energy changes into potential energy. At an extreme position, it has the most potential energy and no kinetic energy. When it moves from this far position to the middle, its potential energy again shifts to kinetic energy. This shows how energy is conserved.
Eventually, the bob stops moving. This occurs because during each swing, some of its energy is moved to the air and used to overcome rubbing at the suspension point. So, the pendulum's energy spreads into the air. Here, the rule of energy conservation is not broken since energy is not ruined but simply changed into a different form.
In simple words: When a pendulum swings, its energy keeps changing between stored (potential) and motion (kinetic). It stops because some energy is lost to air and friction, but this energy isn't gone, just changed form, so the total energy is still conserved.

Exam Tip: Always explain how potential and kinetic energy interconvert at different points of the swing, and account for energy loss to air resistance and friction without violating the conservation law.

Question 16. An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Answer: The starting kinetic energy of the object is \( \frac{1}{2} m v^{2} \). The ending kinetic energy of the object will be \( 0 \). Work done equals the change in kinetic energy.
\( W = E_{K_{f}}-E_{K_{i}}=0-\frac{1}{2} m v^{2} \)
\( W = -\frac{1}{2} m v^{2} \).
The negative sign means that the force is slowing down the object.
In simple words: To stop a moving object, you need to apply work equal to its initial kinetic energy, but in the opposite direction, which is why the work value is negative.

Exam Tip: Remember that work-energy theorem states that the net work done on an object equals its change in kinetic energy. A negative work value indicates that the force opposes the motion.

Question 17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h.
Answer: Given the details: The car's mass, \( m \), is 1500 kg. Its starting speed, \( u \), is 60 km/h. To convert this, \( u = 60 \times \frac {5}{18} = \frac {50}{3} \) m/s.
The car's initial kinetic energy is \( E_{K_{i}} = \frac {1}{2}mu² \).
So, \( E_{K_{i}} = \frac {1}{2} \times 1500 \times \left(\frac{50}{3}\right)^{2} \).
This calculates to \( E_{K_{i}} = 208333.3 \) J.
The car's final kinetic energy, \( E_{K_{f}} \), will be 0 when it stops.
Work done equals the change in kinetic energy.
\( W = E_{K_{f}} - E_{K_{i}} = 0 - 208333.3 \) J.
Therefore, \( W = -208333.3 \) J.
In simple words: To find the work needed to stop a car, first change its speed to meters per second. Then calculate its starting energy. The work required is the negative of this initial energy.

Exam Tip: Always convert units to SI (meters, kilograms, seconds) before calculations in physics problems. The negative sign for work indicates that the work is done against the motion to slow down or stop the object.

Question 18. In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive, or zero.
Answer:

• Work done is zero because the force and displacement are at right angles to each other.
• Work done is positive when the force and displacement go in the same direction.
• Work done is negative when the force and displacement move in opposing directions.

Exam Tip: Remember the formula for work done: \( W = F \times s \times \cos\theta \). If the force and displacement are perpendicular \( (\theta = 90^\circ) \), \( \cos 90^\circ = 0 \), so \( W = 0 \). If they are in the same direction \( (\theta = 0^\circ) \), \( \cos 0^\circ = 1 \), so \( W = Fs \) (positive). If they are in opposite directions \( (\theta = 180^\circ) \), \( \cos 180^\circ = -1 \), so \( W = -Fs \) (negative).

Question 19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer: Yes, Soni is correct. If the total force on a body is zero, then its acceleration will also be zero.
According to Newton's second law, \( F_{net} = ma \).
If the net force \( F_{net} = 0 \),
Then \( 0 = ma \).
This implies that \( a = 0 \).
Therefore, Soni's statement is accurate.
In simple words: Yes, Soni is right because if all the forces pushing and pulling an object cancel each other out, the object won't speed up or slow down, meaning its acceleration is zero.

Exam Tip: Understand Newton's First Law (an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force) and Second Law ( \( F = ma \) ). When net force is zero, acceleration must be zero, even if multiple forces are acting.

Question 20. Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.
Answer: Given the details: Four devices, each with a power of 500 W, are used.
The total time duration, \( t \), is 10 hours.
The energy used by these four devices is calculated as \( W = E = P \times t \).
So, \( W = 4 \times 500 \text{ W} \times 10 \text{ h} = 20000 \text{ Wh} \).
This energy can also be written as \( W = E = 20 \text{ kWh} \), which means 20 units.
In simple words: To find the total energy, multiply the power of one device by the number of devices and then by the hours they are used. Convert the final answer to kilowatt-hours (kWh).

Exam Tip: Remember that 1 kWh (kilowatt-hour) is equivalent to 1 unit of electrical energy. To convert from Watt-hours (Wh) to kilowatt-hours (kWh), divide by 1000.

Question 21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Answer: When an object drops freely towards the ground, its stored energy decreases, and its motion energy increases. As the object touches the ground, all its stored energy changes into motion energy. When the object hits the hard surface, all its motion energy then changes into heat energy and sound energy. It can also reshape the ground, depending on the ground's type and the object's energy.
In simple words: When a falling object hits the ground, its energy of motion doesn't just disappear. Instead, it transforms into other forms like heat and sound energy.

Exam Tip: Always apply the law of conservation of energy. Even if an object appears to lose its mechanical energy, that energy is converted into other forms, such as heat, sound, or deformation energy, but is never destroyed.