by Work, Energy, and Simple Machines
Why does a moving cricket ball knock over wickets? How can a small effort lift a heavy load? And why do two people doing the same work need not have the same power?
The previous chapter, How Forces Affect Motion, explained how forces change motion through Newton’s laws. This chapter takes the next step: understanding work, energy transfer, power, and simple machines that help us perform tasks more conveniently.
What Will Be Learned?
Work Done by a Force
The meaning of work, its SI unit, and positive, negative, and zero work.
Work-Energy Theorem
How work done is connected to a change in an object’s energy.
Forms of Energy
Kinetic energy, potential energy, and conservation of mechanical energy.
Power
How quickly work is done and how to calculate power.
Simple Machines
Fixed and movable pulleys, inclined planes, levers, and mechanical advantage.
15 Textbook Questions
Step-by-step answers that can be opened or hidden for practice.
7.1 Work Done by a Constant Force
In everyday language, work may mean any activity that requires effort. In physics, work is done by a force when the force causes displacement in its direction.
The SI unit of work is the joule (J). Force is measured in newtons (N) and displacement in metres (m).
1 J = 1 N × 1 m = 1 kg·m²/s²Worked Example
A constant force of 50 N moves a box 2 m in the direction of the force. Calculate the work done.
Given: F = 50 N, s = 2 m.
W = F × s W = 50 × 2 = 100 JAnswer: 100 J.
Work from a Force-Displacement Graph
For a constant force, work equals the area under the force-displacement graph. For a varying force, calculate the area under the graph over the relevant displacement interval.
7.1.1 When Is Work Done Zero?
Pushing a Wall
The wall does not move, so its displacement is zero.
W = F × 0 = 0 JCarrying a Bag Horizontally
The supporting force is upward, but the bag’s displacement is horizontal. The supporting force does zero work on the bag.
Work by upward force = 0 J7.1.2 Positive and Negative Work
| Type of work | Condition | Example |
|---|---|---|
| Positive | Force and displacement act in the same direction. | Pushing a wheelchair forward. |
| Negative | Force acts opposite to displacement. | A goalkeeper stopping a moving ball. |
| Zero | No displacement or force perpendicular to displacement. | Pushing an unmoving wall. |
Worked Example: Goalkeeper
A goalkeeper applies a force of 200 N opposite to the displacement of a ball. The ball moves 15 cm while being stopped. Calculate the work done by the goalkeeper on the ball.
s = 15 cm = 0.15 m W = 200 × (−0.15) = −30 JAnswer: −30 J.
7.2 The Work-Energy Theorem
A moving cricket ball can knock over wickets. A raised flowerpot can damage an object if it falls. These examples show that objects can possess a capacity to do work, called energy.
The SI unit of energy is the joule (J), the same as the unit of work.
Positive net work increases kinetic energy; negative net work decreases kinetic energy. Zero net work means no change in kinetic energy.
7.3 Forms of Energy
Energy can appear in different forms and can transform from one form into another.
Mechanical Energy
Energy due to motion or position.
Thermal Energy
Energy associated with the thermal state of matter.
Light Energy
Energy transferred by light and other electromagnetic radiation.
Sound Energy
Energy carried by sound waves.
Electrical Energy
Energy associated with electric charges and circuits.
Chemical and Nuclear Energy
Chemical energy is stored in food and fuels; nuclear energy is stored in atomic nuclei.
7.4 Mechanical Energy
Mechanical energy is the energy an object possesses because of its motion or position.
Mechanical energy = Kinetic energy + Potential energy7.4.1 Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion.
K = ½mv²Here, m is mass in kilograms and v is speed in metres per second. Kinetic energy is measured in joules.
- For the same speed, a larger mass has greater kinetic energy.
- For the same mass, kinetic energy is proportional to the square of speed.
- Doubling speed makes kinetic energy four times as large.
- Tripling speed makes kinetic energy nine times as large.
Worked Numerical: Cricket Ball
A 0.2 kg cricket ball moves at 154.8 km/h. Calculate its kinetic energy.
v = 154.8 × 5/18 = 43 m/s K = ½ × 0.2 × 43² K = 0.1 × 1849 = 184.9 JAnswer: 184.9 J.
Worked Example: Jet Aircraft
A 15,000 kg aircraft is stopped over 100 m by a wire applying an approximately constant backward force of 367,500 N. Calculate its speed before the wire catches the hook.
Work done by the wire is negative because its force is opposite to the displacement.
W = −367500 × 100 = −36,750,000 JThe final kinetic energy is zero, so:
0 − ½mv² = −36,750,000 v² = (2 × 36,750,000) / 15,000 = 4900 v = 70 m/s = 252 km/hAnswer: 70 m/s, or 252 km/h.
7.4.2 Potential Energy
Potential energy is energy stored because of an object’s deformation or the relative positions of objects in a system. A stretched elastic band, a bent bow, and a raised ball are examples.
Here, m is mass, g is acceleration due to gravity, and h is height above the chosen reference level. Near the Earth’s surface, g is about 9.8 m/s²; the textbook sometimes uses 10 m/s² for calculations.
Worked Numerical
A 5 kg object is lifted through a height of 3 m. Find the gain in potential energy using g = 10 m/s².
PE = mgh = 5 × 10 × 3 = 150 JAnswer: 150 J.
7.4.3 Conservation of Mechanical Energy
When only gravity acts and friction and air resistance are neglected, potential energy can change into kinetic energy and vice versa. Their sum remains constant.
KE + PE = ConstantSpeed at the Bottom of a Slide
A child starts from rest at the top of a slide of height h. Ignoring friction and air resistance, the gravitational potential energy at the top becomes kinetic energy at the bottom.
mgh = ½mv² v = √(2gh)In this ideal situation, the final speed depends on the vertical height, not on the child’s mass or the slide’s shape.
Worked Example: Escape Ramp
A 10,000 kg truck moves at 72 km/h and enters a ramp inclined at 30°. Sand exerts a force of 50,000 N opposite to the truck’s motion. Find the minimum ramp length needed to stop it. Take g = 10 m/s² and use the hint that the truck gains 1 m vertically for every 2 m along the ramp.
v = 72 × 5/18 = 20 m/s Initial KE = ½ × 10000 × 20² = 2,000,000 JLet the ramp length be d. The height gained is d/2.
Final PE = mg(d/2) = 50,000d JWork done by the sand is −50,000d J. The change in total energy is 50,000d − 2,000,000 J.
−50,000d = 50,000d − 2,000,000 100,000d = 2,000,000 d = 20 mAnswer: 20 m.
7.5 Power
Power is the rate at which work is done. Two people may do the same amount of work but require different times, so their power outputs can differ.
Average power = Work done / Time taken P = W/tThe SI unit of power is the watt (W).
1 W = 1 J/sOne horsepower is equal to 746 W.
Worked Example: Weightlifter
A weightlifter lifts a 75 kg mass through 2 m in 5 s. Take g = 10 m/s².
W = mgh = 75 × 10 × 2 = 1500 J P = 1500 / 5 = 300 WAnswer: 300 W.
Worked Example: A Car Accelerating
A 1000 kg car starts from rest and reaches 72 km/h in 10 s. Calculate the average power needed to increase its kinetic energy, ignoring losses.
v = 72 × 5/18 = 20 m/s ΔK = ½ × 1000 × 20² = 200,000 J P = 200,000 / 10 = 20,000 W = 20 kWAnswer: 20 kW.
7.6 Simple Machines
Simple machines help perform tasks by changing the magnitude or direction of the force applied. They do not create energy or reduce the total work required in an ideal situation.
The force applied to a machine is called the effort, and the force to be overcome is called the load.
Mechanical advantage (MA) = Load / EffortThe following diagrams are drawn directly in SVG, so no separate image files are required.
7.6.1 Pulley
A pulley is a wheel with a groove that guides a rope. A fixed pulley changes the direction of effort. A movable pulley or a suitable combination of pulleys can lift a heavier load with a smaller effort.
Fixed Pulley
A fixed pulley changes the direction of effort. In the ideal case, the effort and load are equal.
Ideal MA = Load / Effort = 1Movable Pulley
A movable pulley is attached to the load. In the ideal arrangement illustrated, two rope sections support the load, so the effort is half the load.
Ideal MA = 2Compound Pulley
A compound pulley combines fixed and movable pulleys. The ideal mechanical advantage depends on the number of supporting rope sections. The illustrated arrangement has four supporting sections.
Ideal MA = 4 (for the illustrated arrangement)7.6.2 Inclined Plane
An inclined plane is a sloping surface that helps move a heavy load to a higher or lower level. It reduces the effort force by allowing that force to act over a greater distance.
For an ideal inclined plane, assume the load moves at constant speed and friction is neglected. Input work equals the gain in gravitational potential energy.
Effort × Ramp length = Load × Vertical height Ideal MA = L/hSince ramp length L is greater than vertical height h, the ideal mechanical advantage is greater than 1. Making the ramp longer and less steep reduces the effort force but increases the distance over which the effort acts.
Worked Example: Ramp
A ramp has length 50 cm and raises a load to a height of 30 cm. Calculate the ideal mechanical advantage.
MA = L/h = 50/30 ≈ 1.67Answer: 1.67.
7.6.3 Lever
A lever is a rigid bar that rotates about a fixed point called the fulcrum. It has three main parts: effort, load, and fulcrum.
Worked Example: Seesaw
A 15 kg child sits 2 m from the fulcrum. Where should a 30 kg child sit on the other side to balance the seesaw?
15 × 2 = 30 × d 30 = 30d ⇒ d = 1 mAnswer: 1 m from the fulcrum on the opposite side.
Three Classes of Levers
| Class | Arrangement | Examples |
|---|---|---|
| Class I | Fulcrum between effort and load. | Seesaw, scissors, crowbar, pliers, balance scale. |
| Class II | Load between fulcrum and effort. | Lemon squeezer, wheelbarrow, bottle opener. |
| Class III | Effort between fulcrum and load. | Tweezers, broom, oar. |
Interactive Calculator: Work, Energy, Power and Machines
Choose a formula, enter the required values, and calculate the result. For gravitational potential energy, g = 10 m/s² is used.
For work, use displacement in the direction of the force.
Quick Formula Revision
| Concept | Formula | SI unit |
|---|---|---|
| Work | W = F × s, along force direction | J |
| Work-energy theorem | Wnet = ΔK | J |
| Kinetic energy | K = ½mv² | J |
| Gravitational potential energy | PE = mgh | J |
| Mechanical energy | ME = KE + PE | J |
| Average power | P = W/t | W |
| Mechanical advantage | MA = Load/Effort | No unit |
| Ideal inclined plane | MA = L/h | No unit |
| Ideal lever | MA = Effort arm/Load arm | No unit |
Revise, Reflect, Refine — All 15 Textbook Questions
Try each question before showing the solution. Use the controls to reveal all answers or hide them for practice.
Q1. State whether True or False: (i) Work is done when force is applied even if an object does not move. (ii) Lifting a bucket vertically upward results in positive work by the lifting force. (iii) The SI unit for work and energy is joule. (iv) A motionless stretched rubber band has kinetic energy. (v) Energy can change from one form to another.
(i) False. If the object has no displacement, the work done on it by that force is zero.
(ii) True. The lifting force and upward displacement are in the same direction, so the lifting force does positive work.
(iii) True. Both work and energy are measured in joules (J).
(iv) False. A stretched, motionless rubber band stores elastic potential energy, not kinetic energy.
(v) True. Energy can be transformed from one form to another.
Q2. Fill in the blanks: (i) Work done = ____ × ____ in the direction of force. (ii) One joule of work is done when a force of ____ newton displaces an object by 1 metre in the direction of force. (iii) Kinetic energy of a body of mass m and velocity v is ____. (iv) Potential energy of mass m at height h is ____. (v) Power is the ____ at which work is done.
(i) Force × displacement.
(ii) 1 newton.
(iii) K = ½mv².
(iv) PE = mgh.
(v) Rate.
Q3. When a ball thrown upwards reaches its highest point, which statements are correct? (i) The force acting on the ball is zero. (ii) Its acceleration is zero. (iii) Its kinetic energy is zero. (iv) Its potential energy is maximum.
Correct options: (iii) and (iv).
At the highest point, the ball’s instantaneous velocity is zero, so its kinetic energy is zero. Its height and gravitational potential energy are maximum for that flight. Gravity still acts downward, so force and acceleration are not zero.
Q4. Identify the energy transformations: (i) a truck moving uphill, (ii) unwinding of a watch spring, (iii) photosynthesis in green leaves, (iv) water flowing from a dam, (v) burning a matchstick, (vi) explosion of a firecracker, (vii) speaking into a microphone, (viii) a glowing electric bulb, and (ix) a solar panel.
- Truck uphill: Chemical energy from fuel is transformed into mechanical energy; as the truck gains height, gravitational potential energy increases. Some energy also becomes heat and sound.
- Unwinding watch spring: Elastic potential energy transforms into mechanical/kinetic energy.
- Photosynthesis: Light energy becomes chemical energy stored in substances made by the plant.
- Water flowing from a dam: Gravitational potential energy becomes kinetic energy. A turbine can then transform it into electrical energy.
- Burning a matchstick: Chemical energy becomes thermal energy and light.
- Firecracker: Chemical energy becomes light, sound, thermal energy, and kinetic energy of gases and fragments.
- Microphone: Sound vibrations are converted into electrical signals.
- Electric bulb: Electrical energy becomes light and thermal energy.
- Solar panel: Light energy becomes electrical energy.
Q5. A student of mass 50 kg is slowly lifted in an elevator from ground level to the top of a building. The same student later climbs the stairs to the same top. Given h = 72.5 m and g = 10 m/s², find (i) the gain in potential energy when lifted, (ii) the gain when climbing the stairs, and (iii) what this shows about the path taken.
Given: m = 50 kg, h = 72.5 m, g = 10 m/s².
(i) Lifted by elevator:
ΔPE = mgh = 50 × 10 × 72.5 = 36,250 J(ii) Climbing the stairs: The same final height gives the same change in gravitational potential energy.
ΔPE = 36,250 J(iii) Conclusion: The change in gravitational potential energy depends on the vertical height gained, not on the route taken.
Q6. A crane lifts a mass m to the 10th floor in a certain time. It then lifts the same mass to the 20th floor in double the time. How much more energy and power are required? Assume equal height for all floors.
Let each floor have height H and the first time interval be t.
E₁ = mg(10H) = 10mgH P₁ = E₁/t = 10mgH/tFor the 20th floor, both height and time double.
E₂ = mg(20H) = 2E₁ P₂ = E₂/(2t) = P₁Answer: Twice as much energy is required, while the average power remains the same.
Q7. 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 work done? If its speed is doubled, how does the power requirement change?
Energy required: Ideally, the increase in gravitational potential energy is mgh. It depends on the flag’s mass, vertical height, and gravitational acceleration.
Does the speed change the work? Raising the flag to the same height requires the same ideal work, whether slowly or quickly.
What if speed doubles? For the same distance, the time becomes half as long. Since power = work/time, the average power requirement doubles.
Q8. A man of mass 60 kg rides a scooter of mass 100 kg and accelerates it to speed v. The next day his son of mass 40 kg joins him. If the scooter reaches the same speed on both days in the same time, find the ratio of fuel energy used on the two days. Assume all the transferred energy comes from fuel and there are no other losses.
Ignoring losses, the energy supplied equals the increase in kinetic energy.
Day 1: Total mass = 60 + 100 = 160 kg.
E₁ = ½ × 160 × v² = 80v²Day 2: Total mass = 60 + 40 + 100 = 200 kg.
E₂ = ½ × 200 × v² = 100v² E₁ : E₂ = 80v² : 100v² = 4 : 5Answer: Day 1 : Day 2 = 4 : 5.
Q9. On a balanced seesaw, the adult weighs twice as much as the child. Draw a figure showing the distances of the child and the adult from the fulcrum.
For a balanced seesaw, the moments on the two sides are equal.
W × dchild = 2W × dadult dchild = 2dadultAnswer: The child sits twice as far from the fulcrum as the adult.
Q10. A ball of mass 2 kg is thrown upwards with velocity 20 m/s. (i) Identify the sign of work done by gravity during upward and downward motion. (ii) If the ball reaches a height of 19.4 m, how much work is done by air resistance? Take g = 10 m/s².
(i) Work by gravity: Negative during upward motion because gravity acts opposite to displacement. Positive during downward motion because gravity acts in the direction of displacement.
(ii) Work by air resistance:
Initial KE = ½ × 2 × 20² = 400 J Potential energy at 19.4 m = 2 × 10 × 19.4 = 388 JAt maximum height, the ball’s kinetic energy is zero. The mechanical energy has decreased by 12 J.
Wair = 388 − 400 = −12 JAnswer: −12 J. The negative sign means air resistance removes mechanical energy.
Q11. A 10 kg block moves on a horizontal floor with negligible friction. A variable force acts in the direction of motion from 0 m to 4 m as shown in Figure 7.37. Its kinetic energy is 180 J at 0 m. Calculate its speed (i) at 0 m and (ii) at 4 m. Does the block have negative acceleration anywhere?
(i) Speed at 0 m:
K = ½mv² 180 = ½ × 10 × v² ⇒ v² = 36 v = 6 m/s(ii) Speed at 4 m: Work equals the area under the graph.
- Triangle, 0–1 m: ½ × 1 × 50 = 25 J.
- Rectangle, 1–3 m: 2 × 50 = 100 J.
- Triangle, 3–4 m: ½ × 1 × 50 = 25 J.
Negative acceleration? No. The force remains in the direction of motion throughout the graph; it decreases to zero but does not reverse direction.
Q12. Gravity at the Moon’s surface is about one-sixth of gravity on Earth. A ball thrown vertically upward reaches a height of 8 m on Earth. How high will it rise on the Moon if thrown with the same initial velocity?
At the maximum height, initial kinetic energy has converted into gravitational potential energy, neglecting air resistance.
½mv² = mgh ⇒ h = v²/(2g)For the same initial speed, maximum height is inversely proportional to g.
gMoon = gEarth/6 hMoon = 6 × 8 = 48 mAnswer: 48 m.
Q13. A 1000 kg car moves at constant speed. The driver notices an obstruction and applies the brakes. The speed-time graph shows constant speed from A to B and stopping from B to C. (i) Describe the motion from A to B. (ii) Calculate kinetic energy at A. (iii) State the work done by the brakes from B to C. (iv) What does the kinetic energy transform into?
(i) A to B: The car moves at constant speed of 35 m/s.
(ii) Kinetic energy at A:
K = ½mv² = ½ × 1000 × 35² = 612,500 J(iii) Work done by the brakes: The kinetic energy decreases to zero.
W = ΔK = 0 − 612,500 = −612,500 J(iv) Energy transformation: The kinetic energy is mainly transformed into thermal/internal energy through friction in the brakes and tyres, with some energy transferred as sound and deformation.
Q14. The potential-energy-displacement graph of a 0.5 kg ball on a frictionless track shows PE = 30 J at O, 20 J at P, 30 J at Q, and 40 J at R. The ball starts at O from rest. Calculate its speed at P, Q and R.
Given: m = 0.5 kg. At O, the ball is at rest, so total mechanical energy = 30 J.
At P: PE = 20 J.
KE = 30 − 20 = 10 J 10 = ½ × 0.5 × v² ⇒ v² = 40 v = √40 ≈ 6.32 m/sAt Q: PE = 30 J.
KE = 30 − 30 = 0 ⇒ v = 0 m/sAt R: PE = 40 J, which exceeds the total mechanical energy of 30 J. Therefore, the ball cannot reach R from O under the stated initial conditions on a frictionless track.
Answer: P ≈ 6.32 m/s; Q = 0 m/s; R is not reachable under these conditions.
Q15. A coconut of mass 1.5 kg falls from a tree 10 m high into wet sand and comes to rest. (i) Calculate its speed just before impact. (ii) If the average resistive force of the sand is 3000 N and all of the coconut’s energy is used to create the depression, calculate the depth. Take g = 10 m/s².
Given: m = 1.5 kg, h = 10 m, g = 10 m/s², resistive force = 3000 N.
(i) Speed just before impact
Ignoring air resistance, potential energy converts to kinetic energy:
mgh = ½mv² v² = 2gh = 2 × 10 × 10 = 200 v = √200 ≈ 14.14 m/s(ii) Depth of the depression
Initial energy = mgh = 1.5 × 10 × 10 = 150 JUsing work = force × displacement:
150 = 3000 × d d = 150/3000 = 0.05 m = 5 cmAnswers: Speed ≈ 14.14 m/s; depression depth = 5 cm.
Connecting Work and Energy with Force and Motion
In the previous chapter, How Forces Affect Motion, Newton’s laws explained how a net force changes velocity and causes acceleration.
This chapter adds another way to analyse the same events. A force transfers energy when it does work. Positive net work increases kinetic energy, while negative net work reduces it, as when brakes stop a moving vehicle. Potential energy describes energy associated with height or configuration, and conservation of mechanical energy helps predict speed without calculating every instant of motion.
Simple machines connect these principles to everyday life. Pulleys, ramps, and levers can change the effort force or its direction, but they do not create energy.
