NCERT Solutions Class 9 Science Exploration Chapter 07 Work, Energy, and Simple Machines

Question 1. State whether True or False.
(i) Work is said to be done when a force is applied, even if the object does not move. (ii) Lifting a bucket vertically upward results in positive work done on the bucket. (iii) The SI unit for both work and energy is joule (J). (iv) A motionless stretched rubber band has kinetic energy. (v) Energy can change from one form to another.
Answer:
(i) False - Work occurs only when a force creates movement. If nothing moves, no work takes place.
(ii) True - The force applied and the movement are in the same direction, making the work done positive.
(iii) True - Both work and energy are measured in joules (J) according to SI units.
(iv) False - A stationary object contains no kinetic energy. A stretched rubber band holds potential energy instead.
(v) True - Energy shifts between different forms (following the law of conservation of energy).
In simple words: Work needs both force and movement in the same direction. Energy can switch from one type to another, like when a rubber band is stretched and then released.

Exam Tip: Always check three conditions for work: a force exists, the object moves, and movement happens in the force direction. Kinetic energy needs motion - if something is still, it has zero kinetic energy.

Question 2. Fill in the blanks.
(i) Work done = ________________ × ________________ (in the direction of force). (ii) 1 joule of work is done when a force of ________________ newton displaces an object by 1 metre in the direction of the force. (iii) The expression for kinetic energy of a body of mass m and velocity v is ________________. (iv) The potential energy of an object of mass m at a small height h from the Earth's surface is ________________. (v) Power is defined as the ________________ at which work is done.
Answer:
(i) Work done = Force × displacement (in the direction of force).
(ii) 1 joule of work is done when a force of 1 newton displaces an object by 1 metre in the direction of the force.
(iii) The expression for kinetic energy of a body of mass m and velocity v is ½mv².
(iv) The potential energy of an object of mass m at a small height h from the Earth's surface is mgh.
(v) Power is defined as the rate at which work is done.
In simple words: Multiply force by how far it pushes something to get work. Kinetic energy uses the speed squared, and potential energy depends on how high something sits above the ground.

Exam Tip: Remember the formulas: W = F × s, KE = ½mv², PE = mgh. These appear in nearly every problem - write them clearly and show your substitution step by step.

Question 3. When a ball thrown upwards reaches its highest point, tick which of the following statement(s) are correct?
(i) The force acting on the ball is zero. (ii) The acceleration of the ball is zero. (iii) Its kinetic energy is zero. (iv) Its potential energy is maximum.
Answer:
Correct statements: (iii) and (iv)

Explanation:
- Force is not zero - gravity always pulls downward
- Acceleration is not zero - gravity (g) acts downward
- Velocity becomes zero, so KE = 0 ✓
- Height is maximum, so PE is maximum ✓
In simple words: At the top of its flight, the ball stops moving for an instant, giving it zero kinetic energy. Since it reaches its highest point, the potential energy is at its peak. Gravity still acts downward throughout.

Exam Tip: At the highest point, velocity = 0, which means KE = 0. PE peaks because height is greatest. Gravity and acceleration never disappear - they always act on falling/thrown objects.

Question 4. For each of the following situations, identify the energy transformation that takes place:
(i) A truck moving uphill (ii) Unwinding of a watch spring (iii) Photosynthesis in green leaves (iv) Water flowing from a dam (v) Burning of a matchstick (vi) Explosion of a firecracker (vii) Speaking into a microphone (viii) A glowing electric bulb (ix) A solar panel
Answer:
(i) A truck moving uphill - When a truck climbs uphill, it gains height. Its kinetic energy gradually transforms into gravitational potential energy. Energy transformation: Kinetic energy ⇒ Potential energy

(ii) Unwinding of a watch spring - A wound spring carries stored potential energy. As it unwinds, this stored energy produces motion. Energy transformation: Potential energy ⇒ Kinetic energy

(iii) Photosynthesis in green leaves - Plants capture sunlight and use it to make food. Light energy becomes stored chemical energy in the food. Energy transformation: Light energy ⇒ Chemical energy

(iv) Water flowing from a dam - Water sitting at height possesses potential energy. As it flows downward, this energy turns into motion. Energy transformation: Potential energy ⇒ Kinetic energy

(v) Burning of a matchstick - The chemical energy locked in the matchstick releases as heat and light when it burns. Energy transformation: Chemical energy ⇒ Heat energy + Light energy

(vi) Explosion of a firecracker - Firecrackers store chemical energy, which suddenly releases as heat, light, sound, and kinetic energy. Energy transformation: Chemical energy ⇒ Heat + Light + Sound + Kinetic energy

(vii) Speaking into a microphone - When a person speaks, sound energy is created and the microphone converts it to electrical signals. Energy transformation: Sound energy ⇒ Electrical energy

(viii) A glowing electric bulb - Electrical energy entering the bulb becomes light and heat. Energy transformation: Electrical energy ⇒ Light energy + Heat energy

(ix) A solar panel - Solar panels take sunlight and change it straight into electrical energy. Energy transformation: Light energy ⇒ Electrical energy
In simple words: Energy always changes forms. Movement can become height potential, food stores light, and heat plus light come from burning. Machines and nature do these swaps all the time.

Exam Tip: Use the arrow notation (⇒) consistently. Identify what energy goes in and what comes out. For multi-part transformations (like the firecracker), list all outputs separated by plus signs.

Question 5. A student is slowly lifted straight up in an elevator from the ground level to the top floor of a building. Later, the same student climbs the staircase, all the way to the top. Given that the height of the building is h = 72.5 m, acceleration due to gravity is g = 10 m s⁻², and student's mass is m = 50 kg.
(i) Find the gain in the potential energy if the student is lifted straight up to the top. (ii) Find the gain in the potential energy when the student climbs the stairs to the same top. (iii) What do you conclude about the dependence of the potential energy on the path taken?
Answer:
(i) The gain in gravitational potential energy follows the formula: PE = mgh
Substituting values: PE = 50 × 10 × 72.5 = 50 × 725 = 36,250 J
The gain in potential energy = 36,250 J

(ii) The potential energy gain remains the same as part (i), since the final height reached is identical. Therefore, Gain in potential energy = 36,250 J
Explanation: Potential energy relies only on mass, gravity, and height - not on the route taken to reach that height.

(iii) Potential energy shows no dependence on the path. It depends solely on the starting and ending heights. Whether the student travels upward in an elevator or takes the stairs, the potential energy gain stays constant because both arrive at the same height.
In simple words: Taking the elevator or climbing stairs to the same floor stores the same energy in you. How you get there does not matter - only where you end up matters.

Exam Tip: This is a key concept - potential energy is a "state function" and depends only on initial and final positions, not the path. Always use PE = mgh and substitute carefully with the given values.

Question 6. A crane lifts a mass m to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that the height of all floors is equal.
Answer:
Let height of each floor = h
Height to 10th floor = 10h
Height to 20th floor = 20h
Energy = mgh

For 10th floor: E₁ = mg(10h) = 10mgh
For 20th floor: E₂ = mg(20h) = 20mgh

So, E₂ = 2E₁
Energy required is double.

Power = Work / Time
Let time for 10th floor = t
Time for 20th floor = 2t
P₁ = E₁ / t = 10mgh / t
P₂ = E₂ / 2t = 20mgh / 2t = 10mgh / t
So, P₂ = P₁
Power required remains the same.

Summary:
- Energy required is doubled
- Power required remains the same
In simple words: Lifting twice as high takes twice the energy. But if you take twice as long, the power (how fast you work) stays equal because you work more slowly.

Exam Tip: Remember that energy depends on height alone, while power depends on both energy and time. Doubling height means doubling energy, but if time also doubles, power stays constant - this is a common exam pattern.

Question 7. Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at which the flag is raised is doubled, how does the power requirement change? Explain your answers.
Answer:
Factors determining the energy required:
The energy needed to raise the flag equals the gain in its gravitational potential energy. Potential energy (PE) = mgh

Therefore, the factors are:
- Mass of the flag (m)
- Height of the flagpole (h)
- Acceleration due to gravity (g)

Effect of speed on work done:
Raising the flag slowly or quickly does NOT change the amount of work done.

Work done relies only on force and displacement (W = mgh in this case), not on time or speed. Since the flag reaches the same height in both cases, the work done stays the same.

Effect on power when speed is doubled:
Power is defined as: Power = Work / Time
If the speed of raising the flag is doubled, the time taken becomes half.
So, New power = Work / (Time/2) = 2 × (Work/Time)
Therefore, power becomes double.

Conclusion:
- Energy relies on mass, height, and gravity
- Work done stays the same whether the flag is raised slowly or quickly
- If speed is doubled, the power required also doubles
In simple words: Moving fast or slow to the same spot takes the same work. But working faster needs more power because you do the same job in less time.

Exam Tip: Distinguish clearly between work (depends on force and distance only) and power (depends on how fast work is done). This distinction appears frequently in exams and is often tested through comparison questions.

Question 8. A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.
Answer:
Fuel used is proportional to energy required. Energy required = Kinetic Energy (KE) = ½mv²
Since the final velocity is the same on both days, KE depends only on the total mass.

Day 1:
Mass of man = 60 kg
Mass of scooter = 100 kg
Total mass = 60 + 100 = 160 kg
Kinetic energy: KE₁ = ½ × 160 × v²

Day 2:
Mass of man = 60 kg
Mass of son = 40 kg
Mass of scooter = 100 kg
Total mass = 60 + 40 + 100 = 200 kg
Kinetic energy: KE₂ = ½ × 200 × v²

Ratio of fuel used:
Fuel is proportional to KE
KE₁ : KE₂ = 160 : 200 = 4 : 5
The ratio of fuel used on the two days is 4 : 5
In simple words: More mass needs more energy to reach the same speed. The first day uses less fuel because fewer people are on the scooter.

Exam Tip: Always identify what is constant (velocity v) and what changes (total mass). Use KE = ½mv² with the total mass each time, then form the ratio. Cancel out common terms (½ and v²) before simplifying.

Question 9. On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.
Answer:
For a seesaw to be balanced, the clockwise moment must equal the anticlockwise moment.
Moment = Force × Distance from fulcrum

Let: Weight of child = W
Weight of adult = 2W
Distance of child from fulcrum = d₁
Distance of adult from fulcrum = d₂

For balance: W × d₁ = 2W × d₂
Dividing both sides by W: d₁ = 2d₂

Conclusion:
- The child sits at twice the distance from the fulcrum compared to the adult
- The adult sits closer to the fulcrum
- The child sits farther away from the fulcrum

Diagram (representation):

For the seesaw to remain balanced, the lighter child sits farther from the fulcrum and the heavier adult sits closer, such that the child's distance is twice that of the adult.
In simple words: The lighter person must sit farther away to balance the heavier person who sits closer. The distances follow the rule d₁ = 2d₂.

Exam Tip: Apply the principle of moments: effort × effort arm = load × load arm. Always show your moment equation clearly and solve for the distance ratio. A diagram with labeled distances and weights earns marks.

Question 10. A ball of mass 2 kg is thrown up with a velocity of 20 m s⁻¹.
(i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion. (ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume g = 10 m s⁻²)?
Answer:
(i) Upward motion: Work done by gravity is negative (the gravitational force points downward while the ball moves upward, so force opposes motion)
Downward motion: Work done by gravity is positive (the gravitational force and motion both point downward, so force is in the direction of motion)

(ii) Initial KE = ½mv² = ½ × 2 × (20)² = 400 J
Potential energy at height: PE = mgh = 2 × 10 × 19.4 = 388 J

Work done by air resistance = Change in mechanical energy
= Final energy - Initial energy
= 388 - 400
= -12 J
Work done by air resistance = -12 J
In simple words: Gravity pulls down, so it does negative work when the ball rises (working against it) and positive work when the ball falls (helping it). Air resistance always opposes motion and takes away energy.

Exam Tip: Remember that work is negative when force opposes motion. Use the work-energy theorem: the loss in mechanical energy equals the work done against resistance forces like air drag.

Question 11. A 10.0 kg block is moving on a horizontal floor with negligible friction. As shown in Fig. 7.37, a variable force is applied on the block in its direction of motion from its position at 0 m till 4 m. If the block had a kinetic energy of 180 J when it was at 0 m, find the block's speed (i) at 0 m, and (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?

Answer:
Given: Mass = 10 kg, Initial KE = 180 J

(i) Speed at 0 m:
KE = ½mv²
180 = ½ × 10 × v²
v² = 36
v = 6 m/s

Work done = Area under force-displacement graph
From graph:
0-1 m ⇒ triangle = ½ × 1 × 50 = 25 J
1-3 m ⇒ rectangle = 2 × 50 = 100 J
3-4 m ⇒ triangle = ½ × 1 × 50 = 25 J
Total work = 25 + 100 + 25 = 150 J

Final KE = Initial KE + Work = 180 + 150 = 330 J

(ii) Speed at 4 m:
330 = ½ × 10 × v²
v² = 66
v = √66 ≈ 8.12 m/s

Negative acceleration:
No, because the force is always in the direction of motion throughout the entire 0-4 m range. The force never opposes motion, so acceleration is never negative.
In simple words: The area under the force graph tells you total work done. More work means more kinetic energy and higher speed at the end. Since force always pushes forward, there is no backward (negative) acceleration.

Exam Tip: Break the graph area into simple shapes (triangles and rectangles). Add them to find total work, then use work-energy theorem: Final KE = Initial KE + Work. Check whether force ever points backward - if not, acceleration is never negative.

Question 12. The gravitational attraction on the surface of the Moon is about 1/6th of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 m from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?
Answer:
The maximum height reached by a ball thrown upward depends on the acceleration due to gravity.
Using the relation: h = u² / (2g)
For the same initial velocity (u), height is inversely proportional to g: h ∝ 1/g

Given:
Height on Earth = 8 m
Acceleration due to gravity on Moon = (1/6) × g (Earth)

Since gravity on the Moon is 1/6th of that on Earth, the height reached will be 6 times greater.

Height on Moon: hₘ = 6 × 8 ⇒ hₘ = 48 m
The ball will rise up to a height of 48 m on the surface of the Moon. Due to lower gravitational pull on the Moon, the ball experiences less downward acceleration, allowing it to rise to a greater height for the same initial velocity.
In simple words: The Moon pulls down with 1/6 the force. A ball thrown with the same push will go 6 times higher on the Moon than on Earth.

Exam Tip: Recognize the inverse relationship: h ∝ 1/g. If gravity becomes 1/6, height becomes 6 times. Show this proportional reasoning clearly rather than reworking the equation each time.

Question 13. A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig. 7.38.
(i) Describe how the car moves between positions A and B. (ii) Calculate the kinetic energy of the car at A. (iii) State the work done by the brakes in bringing the car to a halt between B and C. (iv) What does the kinetic energy of the car transform into?

Answer:
(i) Between A and B, the car moves with a constant speed of 35 m s⁻¹. This is because the speed-time graph is a horizontal straight line in this interval, which shows that the speed does not change with time. Therefore, the car is in uniform motion and no acceleration acts on it during this part of the motion.

(ii) Given: Mass of the car, m = 1000 kg; Speed at A, v = 35 m s⁻¹
Kinetic Energy = ½mv² = ½ × 1000 × (35)² = 500 × 1225 = 612,500 J
The kinetic energy of the car at A is 6,12,500 J

(iii) When the brakes are applied, the car slows down and finally stops at C. The work done by the brakes equals the change in kinetic energy of the car.
Initial kinetic energy at B = 6,12,500 J
Final kinetic energy at C = 0 J

Work done by brakes = Final KE - Initial KE = 0 - 612,500 = -6,12,500 J
The work done by the brakes is -6,12,500 J. The negative sign shows that the braking force acts opposite to the direction of motion of the car.

(iv) The kinetic energy of the car is mainly transformed into heat energy due to friction between the brake pads and the wheels and also between the tyres and the road. A small part may also be converted into sound energy.
In simple words: The car moves steady until braking. Brakes do negative work because they fight the motion. The energy from moving becomes heat in the brakes and tyres, plus some sound.

Exam Tip: Identify the three regions in a speed-time graph: constant speed (horizontal line), deceleration (sloping line), and stopped (on x-axis). Negative work on an object always removes energy. Always identify where the energy goes - here it becomes heat and sound.

Question 14. The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is 0 m s⁻¹ and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.

Answer:
Mass of the ball, m = 0.5 kg
At point O: Velocity = 0 m s⁻¹, Potential energy = 30 J

Since the track is frictionless, the total mechanical energy of the ball remains constant.

At O: Kinetic energy at O = ½mv² = 0
So, total mechanical energy = Potential energy + Kinetic energy = 30 + 0 = 30 J
Therefore, at every point on the track: Potential energy + Kinetic energy = 30 J

From the graph:
Potential energy at P = 20 J
Potential energy at Q = 30 J
Potential energy at R = 40 J
Now we calculate the velocity at each point.

At P:
Kinetic energy at P = Total energy - Potential energy at P = 30 - 20 = 10 J
Using, KE = ½mv²
10 = ½ × 0.5 × v²
10 = 0.25v²
v² = 40
v = √40 ≈ 6.32 m s⁻¹
So, velocity at P = 6.32 m s⁻¹

At Q:
Kinetic energy at Q = 30 - 30 = 0 J
Thus, v = 0 m s⁻¹
So, velocity at Q = 0 m s⁻¹

At R:
Potential energy at R = 40 J, which is greater than the total mechanical energy (30 J).
This is not possible for the ball because its total energy remains constant at 30 J. Hence, the ball cannot reach point R. So, velocity at R cannot be calculated because the ball never reaches R.

Summary:
Velocity at P = 6.32 m s⁻¹
Velocity at Q = 0 m s⁻¹
The ball cannot reach R
In simple words: At each point, energy stays at 30 J total. When potential energy is low, kinetic energy is high (the ball moves fast). When potential energy is high, kinetic energy drops. The ball cannot go where potential energy exceeds the total.

Exam Tip: Always start by finding total mechanical energy from the given conditions. Then at any point, KE = Total Energy - PE. A point is unreachable if its PE exceeds the total energy.

Question 15. A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand.
(i) Calculate the velocity of the coconut just before it hits the sand. (ii) Assume that the average resistive force of sand is 3000 N and all of the coconut's energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 m s⁻².
Answer:
(i) Given: Mass of coconut, m = 1.5 kg; Height, h = 10 m; Acceleration due to gravity, g = 10 m s⁻²; Initial velocity, u = 0

Using the equation: v² = u² + 2gh
v² = 0 + 2 × 10 × 10 = 200
v = √200 = 10√2 ≈ 14.14 m s⁻¹
The velocity of the coconut just before hitting the sand is 14.14 m s⁻¹

(ii) Just before striking the sand, the coconut has kinetic energy equal to the loss in potential energy during the fall.
Kinetic energy on impact: KE = mgh = 1.5 × 10 × 10 = 150 J
This entire energy is used to do work against the resistive force of sand.
Work done = Force × distance

So, 150 = 3000 × d
d = 150 / 3000
d = 0.05 m
Converting into centimetres: 0.05 m = 5 cm
In simple words: The coconut falls and builds up speed. When it hits sand, all that motion energy pushes it down, making a hole. The deeper you push against strong sand resistance, the more the hole deepens.

Exam Tip: Use v² = u² + 2gh for velocity from free fall. Then convert potential energy (or kinetic energy at impact) to work: W = F × d. Solve for distance d. Always check units and convert to the required form (here, to centimetres).

Exploration Chapter 7 Section-wise Notes

Section 7.1 Work Done by a Constant Force

Scientific definition: Work done = force × displacement in the direction of force
W = F × s

Key conditions for work to be done:

• A force must act on the object
• The object must be displaced
• There must be a component of displacement in the direction of the force

SI Unit: Joule (J)
1 J = 1 N × 1 m = 1 kg m² s⁻²

Work from a force-displacement graph = area under the graph

7.1.1 When is work done equal to zero?

• Force is zero (F = 0)
• Displacement is zero (s = 0), for example, pushing a rigid wall
• Force is perpendicular to displacement, for example, carrying a box while walking horizontally

7.1.2 Positive and Negative Work Done

• Positive work: force and displacement in the same direction, for example, pushing a wheelchair forward
• Negative work: force and displacement in opposite directions, for example, goalkeeper stopping a football

Note: Work does not have a direction; it has only magnitude with a positive or negative sign

Section 7.2 The Work-Energy Theorem

When positive work is done on an object, it gains energy. When negative work is done, it loses energy.

Work-energy theorem: Work done on an object = change in its energy
W = ΔE

This theorem holds for a system of objects and even when forces are not constant.

SI Unit of Energy = Joule (J) - same as work

The joule is named after scientist James Prescott Joule, who studied the relationship between mechanical and thermal energy

Section 7.3 Forms of Energy

Energy can exist in many forms:

Energy can be converted from one form to another. For example:

• Electric bulb: electrical - light + thermal energy
• Ringing bell: mechanical - sound energy
• Food in muscles: chemical - mechanical energy

Section 7.4 Mechanical Energy

Mechanical energy = Kinetic energy + Potential energy

7.4.1 Kinetic Energy

Energy possessed by an object due to its motion.
K = ½mv²

Where m = mass (kg), v = velocity (m/s)

SI Unit: joule (J)

Key points:

• Stationary object: KE = 0
• If velocity doubles, KE becomes 4 times (KE ∝ v²)
• Positive work done - velocity increases - KE increases
• Negative work done - velocity decreases - KE decreases

Derivation: Using W = F × s, F = ma, and v² = u² + 2as:
W = ½mv² - ½mu²
Change in KE = Work done by net force

7.4.2 Potential Energy

Energy stored by an object due to its deformation or due to relative positions in a system.

Types:

• Gravitational potential energy: U = mgh (energy due to height)
• Elastic potential energy: energy stored in stretched/compressed spring, rubber band, bow
• Magnetic potential energy: energy in separated unlike poles
• Electrostatic potential energy: energy in separated charges

Gravitational Potential Energy: U = mgh
Where m = mass (kg), g = 9.8 m/s², h = height (m)

Greater height - greater potential energy

Derived from work-energy theorem: Work done to raise object = mgh = gain in PE

Note: U = mgh is valid only near Earth's surface where g is approximately constant

7.4.3 Conservation of Mechanical Energy

For a freely falling object (no friction):
Mechanical energy = KE + PE = constant = mgh at all points
At any point during fall:

Decrease in PE = Increase in KE
Total mechanical energy remains mgh

This is the Law of Conservation of Mechanical Energy:

"When no external forces other than gravity act on an object, its total mechanical energy remains constant."

Demonstrated by: freely falling object, simple pendulum (bob almost reaches the same height on both sides)

In real life: pendulum slows down due to friction at support and air resistance, some mechanical energy converts to thermal energy

Section 7.5 Power

Power = Rate at which work is done
P = W/t

SI Unit: watt (W), where 1 W = 1 J/s

Other unit: horsepower (hp); 1 hp = 746 W (named after James Watt who invented the efficient steam engine)

Key points:

• More work in same time - more power
• Same work in less time - more power
• Power does not depend on how much work is done, but how fast it is done

Section 7.6 Simple Machines

Simple machines change the magnitude or direction of the force needed to do a task. They do NOT reduce total work done.

Effort: Force applied to the machine
Load: Force to be overcome
Mechanical Advantage (MA) = load/effort

7.6.1 Pulley

• Fixed pulley: changes direction of force only; MA = 1
• Movable pulley: MA > 1; reduces effort needed
• System of pulleys: even greater MA; used in cranes, elevators

7.6.2 Inclined Plane

• Reduces effort needed to raise a load to a height
• MA = L/h (length of inclined plane / height)
• Longer, shallower incline - greater MA - less effort needed
• Trade-off: less force, but applied over greater distance

Derivation: F' × L = mgh - MA = mg/F' = L/h

7.6.3 Lever

A rigid bar rotating about a fixed point (fulcrum)

Three parts: fulcrum, load, effort

Load arm: distance of load from fulcrum
Effort arm: distance of effort from fulcrum

Principle: effort × effort arm = load × load arm
F1 × d1 = F2 × d2
MA = load/effort = effort arm/load arm

Classes of Levers:

Note: A lever reduces force required but not total work done. Effort is smaller but moves through a larger distance.
Machines do not create energy. They only help use energy more effectively

Quick Revision for Exam Day

• W = F × s - know units; identify positive, negative, zero work
• K = ½mv² - KE doubles when v increases √2 times; quadruples when v doubles
• U = mgh - valid near Earth's surface only
• Work-energy theorem: W = change in energy
• Conservation of ME: KE + PE = constant (no friction)
• v = √(2gh) - velocity at base of slide; independent of mass
• P = W/t - 1 watt = 1 J/s; 1 hp = 746 W
• MA = load/effort for all machines
• Inclined plane: MA = L/h
• Lever: effort × effort arm = load × load arm; MA = effort arm/load arm
• Three classes of levers - position of fulcrum, load, effort; examples of each
• Machines do NOT reduce total work - only change force magnitude or direction
• Area under force-displacement graph = work done
• Simple pendulum: PE max at extremes; KE max at centre; total ME constant

Question 1. Define work done by a constant force.
Answer: When a constant force is applied to an object and it moves in the direction of that force, the work done is found by multiplying the force by how far the object travels. The formula is W = F × s. The SI unit for work is the joule (J). An important point is that work itself has no direction - it is a scalar quantity.
In simple words: Work means a force pushes or pulls something, and that thing moves. Multiply the force by the distance it moves, and you get the work done.

Exam Tip: Remember that work requires BOTH force AND displacement in the direction of force. Many students forget this and lose marks by ignoring the direction part.

Question 2. When is work done on an object equal to zero?
Answer: Work becomes zero in three situations: (i) When there is no force applied at all, (ii) When the object does not move even though a force is present - for example, when you push against a solid wall and nothing budges, or (iii) When the force acts at a right angle to the direction the object moves - like carrying a box while walking straight ahead, where the upward support force is perpendicular to the forward motion.
In simple words: Work is zero if there is no force, no movement, or if the force and movement are at right angles to each other.

Exam Tip: The perpendicular case often confuses students. Always check the angle between force and displacement before deciding if work is zero.

Question 3. Define 1 joule of work.
Answer: One joule is the amount of work performed when a force of 1 newton causes an object to move 1 metre in the same direction as the force. In equation form: 1 J = 1 N × 1 m. This can also be written in terms of mass, length, and time: 1 J = 1 kg m² s⁻².
In simple words: One joule happens when you push with 1 newton of force and something moves 1 metre forward.

Exam Tip: Know the alternate form 1 kg m² s⁻² - examiners sometimes ask for this SI base unit expansion to test dimensional analysis skills.

Question 4. What is kinetic energy? Give its formula.
Answer: Kinetic energy is the energy that a moving object has because of its motion. Any object in motion carries this form of energy. The formula for kinetic energy is K = ½mv², where m represents the mass of the object in kilograms and v is the speed of the object in metres per second. The unit of kinetic energy, like all energy, is the joule (J).
In simple words: When something moves, it has kinetic energy. The faster or heavier it is, the more kinetic energy it has.

Exam Tip: Note that kinetic energy depends on the SQUARE of velocity - double the speed means four times the energy. This is a high-yield concept for exams.

Question 5. What is gravitational potential energy? Give its formula.
Answer: Gravitational potential energy is energy that gets stored in the system made up of Earth and an object sitting above the ground. This stored energy depends on how high the object is. The formula is U = mgh, where m is the object's mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height above ground level.
In simple words: A heavy thing up high has more potential energy than the same thing down low, because it can fall and do more damage.

Exam Tip: Always use g = 9.8 m/s² unless the question says otherwise. And remember that height h is always measured from a reference point - usually the ground.

Question 6. State the work-energy theorem.
Answer: The work-energy theorem says that the amount of work done on an object equals how much its total energy changes. Expressed as an equation: W = ΔE. When work done is positive (force in the direction of motion), the object's energy goes up. When work done is negative (force opposite to motion), the object's energy comes down. This principle is true not just for constant forces, but also for forces that vary during the motion.
In simple words: Work done on something changes how much energy that thing has. More work means more energy.

Exam Tip: This theorem is the bridge between forces and energy. Use it when questions jump between force-based and energy-based descriptions of the same situation.

Question 7. Define power and give its SI unit.
Answer: Power is defined as how fast work gets done - it is the rate at which energy is being transferred. The formula is P = W/t, where W is the work and t is the time taken. The SI unit of power is the watt (W). One watt equals one joule per second (1 W = 1 J/s). The unit is named to honour James Watt, an engineer who made the steam engine much more useful.
In simple words: Power is how quickly you do work. Doing the same work in half the time means twice the power.

Exam Tip: Don't confuse power (watts) with energy (joules). Power is a rate - it has time in the denominator.

Question 8. What is mechanical energy?
Answer: Mechanical energy is the sum of all the kinetic energy and potential energy that an object possesses at any given moment. It is written as ME = K + U, which expands to ME = ½mv² + mgh. For an object that is falling freely and has no friction or air resistance acting on it, the total mechanical energy stays exactly the same throughout the fall - it never increases or decreases.
In simple words: Mechanical energy is kinetic plus potential energy added together. For a falling object with no friction, this total never changes.

Exam Tip: This is the basis for conservation of mechanical energy. When solving free-fall problems, set mechanical energy at the top equal to mechanical energy at the bottom.

Question 9. What is a simple machine?
Answer: A simple machine is a basic tool that makes work easier for us. It does this by changing either how much force we need to use, or the direction in which we need to push or pull. However, an important fact is that simple machines do not reduce the actual amount of work - they only redistribute that work across force and distance differently. Everyday examples include pulleys, inclined planes, and levers.
In simple words: A simple machine lets you use less force, but you have to move through a greater distance. The total work stays the same.

Exam Tip: Students often think machines reduce total work - they don't. Machines trade force for distance. This is a common exam trap.

Question 10. Define mechanical advantage.
Answer: Mechanical advantage is a number that compares the load (the weight or force we want to move) to the effort (the force we actually apply). The formula is MA = load/effort. When the mechanical advantage is greater than 1, it means the machine is amplifying your applied force - you can lift something heavier than the force you put in. A mechanical advantage of exactly 1 means the machine provides no force multiplication, only a change in direction.
In simple words: Mechanical advantage tells you how much the machine multiplies your effort. Higher numbers mean the machine helps you more.

Exam Tip: Always express MA as a simple number without units. If MA = 3, you can lift 3 times heavier a load with the same effort.

Question 11. What is the mechanical advantage of a fixed pulley?
Answer: A fixed pulley is one that is attached to a support and does not move. Its main role is to change the direction of the applied force - it lets you pull downward instead of having to lift upward. However, the size of the force you must use stays the same - the pulley does not reduce the force needed. Because the effort you provide equals the load you want to lift, the mechanical advantage of a fixed pulley is always exactly 1.
In simple words: A fixed pulley just changes which way you pull. It does not make lifting easier, only more convenient.

Exam Tip: Don't confuse fixed and movable pulleys. Remember: fixed pulley has MA = 1; movable pulley has MA > 1.

Question 12. Write the formula for mechanical advantage of an inclined plane.
Answer: The mechanical advantage of an inclined plane is given by the formula MA = L/h, where L is the length measured along the sloped surface and h is the vertical height that needs to be gained. This formula shows that as the inclined plane becomes longer and shallower for the same height, the mechanical advantage increases. This means less force is required to push an object up the slope.
In simple words: The longer and gentler the slope, the easier it is to push something up. A very steep short slope needs more force.

Exam Tip: Always measure L along the incline itself and h as a vertical rise. A common mistake is measuring L horizontally instead.

Question 13. What is the principle of a lever?
Answer: The lever principle describes how a lever works in balance. It states that the effort multiplied by its distance from the pivot point equals the load multiplied by its distance from the pivot point. Written as an equation: effort × effort arm = load × load arm, or F1 × d1 = F2 × d2. By making the effort arm longer than the load arm, you can apply a smaller effort to move a much larger load. The mechanical advantage of a lever is found by dividing the effort arm length by the load arm length: MA = effort arm/load arm.
In simple words: The farther from the pivot you push, the less force you need to lift something heavy on the other side.

Exam Tip: The pivot point (fulcrum) is key. In problems, always identify it first, then measure arms from there.

Question 14. If velocity of an object doubles, how does its kinetic energy change?
Answer: Using the kinetic energy formula K = ½mv², if the velocity doubles from v to 2v, the new kinetic energy becomes KE = ½m(2v)² = ½m × 4v² = 4 × (½mv²). This means the kinetic energy becomes 4 times the original value. The reason is that kinetic energy is proportional to the square of velocity - so doubling the velocity causes the kinetic energy to increase by a factor of 4 (which is 2 squared).
In simple words: Double the speed, and you quadruple the kinetic energy. This is why fast-moving things are so much more dangerous than slow ones.

Exam Tip: This quadrupling relationship explains why stopping a car from 80 km/h takes much more braking distance than from 40 km/h. Examiners love this concept.

Question 15. Name the three classes of levers with one example each.
Answer: The three classes of levers are arranged according to where the fulcrum, load, and effort sit relative to each other.

Class I - Fulcrum between load and effort: The fulcrum sits in the middle, with the load on one side and the effort on the other side. Examples include a seesaw and scissors.

Class II - Load between fulcrum and effort: The load is in the middle position, with the fulcrum at one end and the effort at the other end. Examples include a wheelbarrow and a bottle opener.

Class III - Effort between fulcrum and load: The effort is in the middle, with the fulcrum on one side and the load on the other side. Examples include tweezers and a broom.
In simple words: Class I has the pivot in the middle (like a seesaw). Class II has the load in the middle (like a wheelbarrow). Class III has the effort in the middle (like tweezers).

Exam Tip: A quick way to remember is the positions: Class I (F-L-E order), Class II (F-L-E arrangement), Class III (F-E-L). Practice identifying the fulcrum first in any lever diagram.

Question 16. Explain why a person pushing a rigid wall does no work on the wall, yet feels tired.
Answer: To do scientific work, two things must happen at the same time: a force must be applied AND the object must move in the direction of that force. Even though the person pushes on the wall with a lot of force, the wall does not budge - the displacement is zero. Using the work formula W = F × s = F × 0 = 0, we get zero work. However, inside the person's body, muscles are constantly tightening and relaxing, which burns up chemical energy stored in the body. This energy use inside the muscles causes the feeling of tiredness, even though the work done on the external wall is zero.
In simple words: Work needs both force AND movement. The wall does not move, so no work happens. But your muscles use energy trying, which is why you feel tired.

Exam Tip: This question tests understanding that work requires BOTH components. Don't just look at force alone.

Question 17. Why does a goalkeeper do negative work on the football while stopping it?
Answer: When a goalkeeper catches or blocks a moving ball, the force applied by the goalkeeper's hands or body points backward - opposite to the direction the ball is moving forward. Since the force and the displacement are in opposite directions (the ball moves forward but the force points backward), the work done is negative according to W = F × (-s) = negative. This negative work is important because it removes energy from the ball, slowing it down from its initial speed until it stops completely at zero velocity.
In simple words: Negative work removes energy. When the goalkeeper pushes backward against a forward-moving ball, that is negative work, and it slows the ball down.

Exam Tip: Negative work always opposes motion and removes energy from an object. Use this idea whenever a force slows something down.

Question 18. Distinguish between kinetic energy and potential energy with one example each.
Answer: Kinetic Energy is energy that an object has due to being in motion. It is found using the formula K = ½mv². A practical example: a cricket ball rolling or flying through the air has kinetic energy and can knock over the wickets because of this moving energy. Potential Energy is energy that is stored because of an object's location or the shape it has been stretched or compressed into. For gravitational potential energy, the formula is U = mgh. A practical example: a flowerpot sitting on a high shelf stores gravitational potential energy; when it falls, this stored energy converts into kinetic energy. Both types of energy are measured in joules (J).
In simple words: Kinetic energy is from motion. Potential energy is from position - something ready to fall or move.

Exam Tip: Remember that potential energy can convert to kinetic energy and vice versa. This is the foundation of conservation of mechanical energy.

Question 19. A child of mass m slides down a frictionless slide of height h. What is the velocity at the bottom? Does it depend on the mass or shape of the slide?
Answer: At the top of the slide: PE = mgh and KE = 0. At the bottom: PE = 0 and KE = ½mv². Applying conservation of mechanical energy: mgh = ½mv². Dividing both sides by m, we get gh = ½v², which simplifies to v = √(2gh). Notice that the mass m cancels out completely, so the velocity at the bottom depends only on the height h, not on the child's mass. The shape of the slide also makes no difference as long as the total height and the absence of friction remain the same. A winding slide and a straight slide would result in the same final velocity if they have the same height drop.
In simple words: The speed at the bottom depends only on how high up you started, not on how heavy you are or what path you took.

Exam Tip: The fact that mass cancels is surprising to many students but very important. This makes these problems much simpler than they first appear.

Question 20. Explain conservation of mechanical energy using a simple pendulum.
Answer: Consider a pendulum swinging back and forth. At the highest point on the left side (position P), the bob momentarily stops, so all its energy is potential energy (mgh) and its kinetic energy is zero. As the bob swings down to the lowest point at the middle (position Q), the potential energy converts entirely into kinetic energy - now PE is zero and KE is maximum. When the bob swings up to the highest point on the right side (position R), the kinetic energy converts back into potential energy again. Throughout this entire motion, the sum of KE and PE (the total mechanical energy, which equals mgh) stays constant, proving the law of conservation of mechanical energy. In real life, however, friction at the support point and air resistance gradually take away some of this mechanical energy, which is why the pendulum eventually comes to rest.
In simple words: A pendulum swaps kinetic and potential energy back and forth, but the total stays the same. Friction eventually stops it.

Exam Tip: Always identify the points of maximum PE (highest points) and maximum KE (lowest point) when answering about pendulums. Use this to set up energy conservation equations quickly.

Question 21. How does an inclined plane act as a simple machine? Why does a longer, shallower incline require less effort?
Answer: An inclined plane helps by allowing you to spread out the work of raising a heavy object over a longer distance instead of lifting it straight up. Using the work-energy theorem, the work input equals the work output: F' × L = mgh, where F' is the effort (force) needed, L is the length of the slope, and mgh is the gravitational potential energy the object gains. Rearranging gives F' = mgh/L. The mechanical advantage is MA = L/h. As you make the inclined plane longer (larger L) for the same height (h), the MA gets bigger, meaning the force F' required becomes smaller proportionally. The total work done (mgh) remains exactly the same no matter whether you lift straight up or use the incline - the machine does not reduce work, but it redistributes the force across a longer distance, making the effort at any single moment much easier on the muscles.
In simple words: A longer, gentler slope lets you spread the work over more distance, so you do not have to push as hard at any moment.

Exam Tip: Always emphasize that total work stays the same. The benefit is that less force is needed at once, not that total effort disappears.

Question 22. State the three classes of levers with their characteristic feature and two real-life examples of each.
Answer: Class I Lever - The fulcrum lies between the load and the effort. The effort and load are on opposite sides of the fulcrum. The mechanical advantage can be greater than, equal to, or less than 1 depending on the arm lengths. Real-life examples: a seesaw, scissors, a crowbar, pliers, and a balance scale.

Class II Lever - The load lies between the fulcrum and the effort. The effort arm is always longer than the load arm, which means the mechanical advantage is always greater than 1. This makes these levers particularly good for lifting or moving heavy objects. Real-life examples: a wheelbarrow and a bottle opener, a lemon squeezer.

Class III Lever - The effort lies between the fulcrum and the load. The load arm is always longer than the effort arm, which means the mechanical advantage is always less than 1. Although these levers don't provide a force advantage, they let you move the load through a larger range and at a higher speed. Real-life examples: tweezers, a broom, a hammer, and an oar.
In simple words: Class I has the pivot in the center. Class II gives you a force boost. Class III gives you speed and range instead of power.

Exam Tip: Draw simple diagrams for each class showing the positions of F, L, and the fulcrum. This visual representation will help you identify any lever type in exam diagrams.

Question 23. A man runs up a flight of stairs in 10 seconds. His friend walks up the same stairs in 50 seconds. Who does more work and who has more power?
Answer: Assume both the man and his friend have the same body mass m and climb the same vertical height h of the stairs. Since work equals mass times gravitational acceleration times height (W = mgh), both individuals do exactly the same amount of work - neither does more work than the other. However, power is defined as the rate of doing work: P = W/t. The man completes the same work in 10 seconds, so his power is W/10. His friend takes 50 seconds for the same work, so his friend's power is W/50. Comparing these: the man's power is 5 times larger than his friend's power (W/10 is 5 times W/50), even though they both did equal work. This shows an important distinction: two people can do the same total work but have very different power outputs depending on how fast they work.
In simple words: Both do the same work. The man does it faster, so he has more power. Power is about speed, not total effort.

Exam Tip: This question teaches that work and power are different. Same work, different times = different powers. Use this in real-world scenarios like comparing athletes or machines.

Question 24. Explain energy transformations in a watermill (gharat or panchakki) as described in the chapter.
Answer: In a traditional Himalayan watermill (called a gharat or panchakki), water is held at an elevated location and possesses gravitational potential energy by virtue of its height. As this water is released and flows downhill through a pipe or channel (labeled A), the potential energy gets converted into kinetic energy because the water gains speed while falling. This fast-moving water strikes a rotating wheel (labeled B) with force, transferring its kinetic energy to the wheel and causing it to spin. The rotational kinetic energy of the spinning wheel is then passed along a shaft that connects to a grinding stone (labeled C) positioned above. The rotational motion drives the grinding stone, which converts the rotational energy into mechanical energy for the practical task of grinding grain into flour. Modern hydroelectric power plants work on this identical principle - water stored behind a dam has gravitational PE that converts to KE as it falls, and this moving water drives turbines that generate electrical energy instead of grinding grain.
In simple words: High water has potential energy. Falling water gains speed and becomes kinetic energy. Moving water turns a wheel that grinds grain.

Exam Tip: Trace the energy transformations step by step: PE to KE to rotational motion to mechanical work. This chain of conversions is a complete answer.

Question 25. Why do roads on hills wind around in gentle slopes rather than going straight up? Explain using the concept of inclined planes.
Answer: A road that goes straight up a hill vertically would force a vehicle to apply a force equal to its own full weight (F = mg) - an enormous effort. A winding road that loops around the hillside acts as a very long inclined plane with a gentle slope. By increasing the path length L while the vertical height h stays constant, the mechanical advantage (MA = L/h) becomes much larger. This larger mechanical advantage means the engine has to supply only a much smaller force to climb the same height. The total work done (W = mgh) remains identical whether you go straight up or take the winding route, but the winding path spreads this work across a much greater distance. As a result, the force needed at any one point is much smaller and manageable, making the climb achievable for regular vehicles instead of requiring extremely powerful engines.
In simple words: Winding roads are longer but much gentler. A gentle slope lets vehicles climb with less effort because the long distance spreads out the work.

Exam Tip: This is a real-world application of the inclined plane concept. Mention that the work is the same, but the force is much reduced - that shows full understanding.

Question 26. Define work done. When is work done zero? Distinguish between positive and negative work with one example each.
Answer: Work done by a constant force acting on an object is calculated as the product of the force and the displacement occurring in the direction of the force, given by the formula W = F × s, with the SI unit being joule (J). Work becomes zero in three different situations: (i) when the force applied is zero, (ii) when the object does not move (zero displacement) despite force being applied, such as when pushing against a rigid wall, or (iii) when the force is applied perpendicular to the direction of displacement, like carrying a box horizontally while walking forward.

Positive work occurs when force and displacement point in the same direction. Example: pushing a wheelchair forward - both the applied force and the resulting motion point in the same direction, so the work done by the person is positive. Negative work occurs when force and displacement point in opposite directions. Example: a goalkeeper stopping a football - the force applied by the goalkeeper points backward, opposite to the ball's forward motion, so the work done is negative.
In simple words: Work is force times distance. Work is zero if nothing moves. Work is positive when force helps motion; negative when it blocks motion.

Exam Tip: Always check the direction relationship between force and displacement. This determines the sign of work - it is the most commonly missed detail.

Question 27. Derive the expression for kinetic energy using the work-energy theorem. What happens to kinetic energy when velocity doubles?
Answer: Consider an object with mass m beginning from rest (initial velocity u = 0) that is acted upon by a constant force F. The object travels a displacement s and reaches a final velocity v. Using kinematic equations: v² = u² + 2as, which simplifies to v² = 0 + 2as, giving us s = v²/(2a). The work done is W = F × s = ma × v²/(2a) = ½mv². According to the work-energy theorem, the work done equals the change in kinetic energy, so the kinetic energy is K = ½mv².

If the velocity doubles from v to 2v: The new kinetic energy becomes KE = ½m(2v)² = ½m × 4v² = 4 × (½mv²). This shows that kinetic energy quadruples when velocity doubles. The relationship K ∝ v² means kinetic energy is proportional to the square of velocity. This explains a vital real-world fact: doubling the speed of a vehicle makes it four times harder to stop because the brakes must perform four times more work to remove four times the kinetic energy.
In simple words: Double the speed, and kinetic energy becomes four times bigger. That is why fast crashes are so much more damaging than slow ones.

Exam Tip: Understand the derivation - don't just memorize the formula. Examiners often ask you to show steps, and the logic will help you answer follow-up questions.

Question 28. State and explain the Law of Conservation of Mechanical Energy for a freely falling object. Show that mechanical energy remains constant at every point.
Answer: The Law of Conservation of Mechanical Energy states: When only gravity acts on an object (no air resistance or friction), the total mechanical energy, which is the sum of kinetic and potential energy, stays constant throughout the motion.

For an object of mass m dropped from height h, examine different points:

At Point A (the top): KE = 0 (starts from rest), PE = mgh (at maximum height), Total ME = 0 + mgh = mgh

After falling for time t, reaching Point B at height h': PE = mgh' = mgh - ½mgt² and KE = ½mv² = ½mgt² (from v = gt). Total ME = (mgh - ½mgt²) + ½mgt² = mgh

At ground level (h = 0): KE = mgh (all PE converted), PE = 0 (no height left), Total ME = mgh + 0 = mgh

At every single point during the fall, the total mechanical energy equals mgh - it remains constant. The potential energy that decreases exactly equals the kinetic energy that increases, proving conservation of mechanical energy. Energy is not lost; it is simply transformed from one form to another.
In simple words: As something falls, it loses height (less PE) but gains speed (more KE). The total energy stays the same - energy just changes form.

Exam Tip: Show the calculation at least three points (top, middle, bottom) to fully prove conservation. Examiners specifically look for this completeness.

Question 29. Explain the three simple machines studied in Chapter 7 - pulley, inclined plane, and lever. Give the formula for mechanical advantage of each.
Answer: A pulley is a wheel with a grooved rim that guides a rope or cable around it. A fixed pulley works mainly to change the direction of the force - it allows you to pull downward instead of pulling upward, which is more convenient. Since the effort you apply equals the load you lift, the mechanical advantage is MA = 1. Movable pulleys, however, have MA > 1 because they can lift heavier loads while you apply less effort.

An inclined plane is a sloped surface that assists in raising heavy objects to a height by distributing the work over a longer distance. Using the work-energy theorem: the work input F' × L equals the work output mgh, giving F' = mgh/L. Therefore, MA = L/h. A longer, shallower incline produces a greater mechanical advantage, meaning less force is needed to raise the object, but you must travel a greater distance. The total work remains constant.

A lever is a rigid bar that rotates about a support point (fulcrum). The operating principle states: effort × effort arm = load × load arm. The mechanical advantage is MA = effort arm/load arm. When the effort arm is longer than the load arm, a smaller effort can overcome a larger load. Levers are divided into three classes based on the positions of the effort, load, and fulcrum relative to each other. In all cases, machines make work easier by reducing force but they never reduce the total work done - they trade force for distance or distance for force.
In simple words: All three machines help you use less force, but you move farther. They make work easier without making it smaller.

Exam Tip: Remember the key formulas: pulley MA = 1 (fixed), inclined plane MA = L/h, lever MA = effort arm/load arm. These are high-frequency exam questions.

Question 30. Define power. Calculate power in two practical examples from the chapter. Why is the unit watt named after James Watt?
Answer: Power is defined as the rate at which work is performed - how quickly energy is transferred or used. The formula is P = W/t, where W is the work done and t is the time taken. The SI unit is the watt (W), where 1 watt equals 1 joule per second (1 W = 1 J/s). The unit honours James Watt, an engineer who developed a highly efficient steam engine capable of producing continuous rotational motion - a breakthrough that powered the Industrial Revolution and transformed manufacturing and transportation.

Example 1 - A Weightlifter: A weightlifter raises a 75 kg mass vertically by 2 metres in 5 seconds. Work done = mgh = 75 × 10 × 2 = 1500 J. Power = W/t = 1500 J / 5 s = 300 W.

Example 2 - A Car Engine: A 1000 kg automobile accelerates from rest to 20 m/s in 10 seconds. Work done equals the kinetic energy gained: W = ½mv² = ½ × 1000 × (20)² = ½ × 1000 × 400 = 200,000 J. Power = W/t = 200,000 J / 10 s = 20,000 W = 20 kW.

These examples show a key distinction: two workers performing the same work but taking different amounts of time will have the same total energy use but different power outputs. Power measures how fast energy is transferred, not the total amount of energy transferred.
In simple words: Power is work done divided by time taken. The faster you do work, the more power you use. James Watt's steam engine was powerful, so the unit is named after him.

Exam Tip: Always show both work calculation and time division separately - this shows complete understanding and prevents arithmetic errors.

Important Formulae - Class 9 Science Exploration Chapter 7