Work, Energy, and Simple Machines gives you an energy method for analysing motion. It defines work done by a constant force as force times displacement along the force, links work to energy through the work-energy theorem, and derives kinetic energy ½mv² and potential energy mgh. It shows that mechanical energy stays constant when only gravity acts, defines power as the rate of doing work, and explains how pulleys, inclined planes and levers make tasks easier without reducing the total work. Core skills are finding the sign of work and solving energy, power and mechanical advantage numericals.
Key Concepts
1. Work Done by a Constant Force
In science, “work” has one exact meaning. NCERT builds it from lifting 5 kg bags of wheat slowly through 1 m with an upward force equal to the weight, mg.
- Lift 3 bags together through 1 m: 3 times the force, 3 times the work. So a larger force over the same distance does proportionally more work.
- Lift one bag through 3 m: 3 times the distance, 3 times the work. So the same force over a larger distance does proportionally more work.
These observations give the scientific definition (Eq. 7.1):
Work done on an object by a constant force = force applied × displacement in the direction of the force
W = F × s (Eq. 7.2)
The rule works in any direction, as long as s is the displacement along the force. Always name the force (or agency) doing the work and the object on which it is done.
The joule
The SI unit of work is the joule (J). Since W = F × s, 1 J = 1 N × 1 m. 1 joule of work is done on an object when a constant force of 1 newton is applied to it and it is displaced by 1 metre in the direction of the force. Because 1 N = 1 kg m s−2, 1 J = 1 kg m2 s−2.
Force and displacement both have magnitude and direction. Work has no direction. It is described by a number with a positive or negative sign.
Work as the area under a force-displacement graph
For a constant 10 N force over 1 m, the force-displacement graph is a horizontal line and the rectangle under it has area 10 N × 1 m = 10 J, the work done. Even when the force is not constant, the work done equals the area under the force-displacement graph between the initial and final positions.
When is work done equal to zero?
| Condition | Why W = 0 | Example |
|---|---|---|
| No force, F = 0 | F × s = 0 × s = 0 | No force acts on the object at all |
| No displacement, s = 0 | F × 0 = 0, however large the force | Pushing a rigid wall; a weightlifter holding a barbell steady |
| Force perpendicular to displacement | No displacement in the direction of the force | A girl carrying a box while walking: her upward force balances its weight, the box moves horizontally |
You feel tired pushing a wall because your muscles repeatedly expand and contract and use up your body’s internal energy, yet you do no work on the wall.
Positive and negative work
- Positive work: the displacement is in the same direction as the force. Pushing a wheelchair forward: you do positive work on it.
- Negative work: the displacement is opposite to the force. A goalkeeper stopping a football pushes against its motion: she does negative work on the ball. In the formula, take the displacement as negative.
The ball and goalkeeper push on each other with equal and opposite forces: the ball does positive work on the goalkeeper, the goalkeeper negative work on the ball. Friction on a sliding object acts against its motion, so it does negative work.
Example 7.1 (NCERT): a girl lifts a dumbbell and slowly lowers it. While lifting, her force (equal to its weight, upward) is along the displacement: positive work. While lowering, she still pushes up to hold it, but it moves down: negative work.
2. Energy and the Work-Energy Theorem
A thrown cricket ball knocks over the wicket; a raised flowerpot can damage something below if it falls. Each has a capacity to do work. An object having the capacity to do work is said to possess energy. The ball got its energy from the work done by the fielder, the pot from the work done in raising it. Positive work on an object gives it energy, which it can pass on by pushing another object, as the ball does to the wickets.
Work-energy theorem: work done on an object = change in its energy (Eq. 7.3)
- It also holds for a system of objects and when the forces are not constant.
- Positive work increases the energy; negative work decreases it.
- The SI unit of energy is the joule, named after James Prescott Joule, who studied how mechanical and thermal energy convert into each other.
Carrom (Example 7.3): the striker hits the white coin, which hits the black coin. The striker does positive work on the white coin (its energy rises); by Newton’s third law the white coin does negative work on the striker (its energy falls). Likewise the white coin does positive work on the black coin and the black coin negative work on the white coin.
Forms of energy
Energy exists in many forms, and it can change from one form to another.
| Form | What it is (NCERT Fig. 7.10) |
|---|---|
| Mechanical energy | Energy due to motion or position of objects |
| Thermal energy | Energy that makes things warm or hot |
| Light energy | Energy that allows us to see |
| Sound energy | Energy of vibrations of air or other molecules |
| Electrical energy | Energy related to position or motion of charges |
| Nuclear energy | Energy stored in the nuclei of atoms |
| Chemical energy | Energy stored in fuels and food in the form of chemical bonds between atoms |
Conversions: bulb, electrical to light; water heater, electrical to thermal; food, chemical to mechanical in muscles; ringing bell, mechanical to sound. Pedalling a bicycle turns muscular energy into kinetic energy, and friction and air resistance turn part of it into heat.
3. Kinetic Energy
Mechanical energy is the energy an object possesses due to its motion or position. Its two parts are kinetic energy and potential energy.
Kinetic energy is the energy possessed by an object due to its motion. A moving bicycle and a rolling ball have it. An object at rest is defined to have zero kinetic energy.
Deriving the expression
Take an object of mass m moving with initial velocity u. A constant force F acts on it along its motion, so its acceleration a is constant, and after a displacement s its velocity is v.
- Kinematic equation: v2 = u2 + 2as, so s = (v2 − u2) / (2a) (Eq. 7.4)
- Work done: W = F × s, and Newton’s second law gives F = ma, so W = ma × s.
- Substitute s: W = ma × (v2 − u2) / (2a)
- The a cancels: W = ½m(v2 − u2) (Eq. 7.5)
By the work-energy theorem, this work is the change in the object’s energy. If the object starts from rest (u = 0), the whole change is its final kinetic energy:
K = ½mv2 (Eq. 7.6)
- SI unit: joule. Kinetic energy has no direction and is never negative.
- Positive work on the object: velocity and kinetic energy increase.
- Negative work: velocity and kinetic energy decrease.
- No work (W = 0) and no change of velocity: kinetic energy stays constant. An object moving with constant velocity has the same kinetic energy at every position.
How K depends on mass and speed
| Change | Effect on K = ½mv2 |
|---|---|
| Mass doubled, same speed | K doubles (K ∝ m) |
| Speed doubled, same mass | K becomes 4 times (K ∝ v2), NCERT Example 7.4 |
4. Potential Energy
Release a stretched slingshot (gulel) band or a bent bow and the stone or arrow flies off with kinetic energy it did not have. That energy came from the band or bow, which had a capacity to do work due to its shape: the work done to deform it was stored in it. A stretched or compressed spring stores energy the same way and, on release, returns to its shape and hands the energy to the object in contact as kinetic energy.
Energy stored by relative position
- Magnets: unlike poles pulled apart and released rush together with kinetic energy (strong magnets can crush a pea between them).
- Electric charges: separated charges also store energy that can do work when released.
- Earth and ball: lifting the ball needs work against gravity; released, ball and Earth rush together and gain kinetic energy.
Definition: the energy stored by an object as a result of its deformation, or in a system of objects due to their relative positions, is called potential energy.
Gravitational potential energy
The Earth hardly moves towards a ball, so the Earth-ball system’s stored energy is called the gravitational potential energy of the ball. In this chapter “potential energy” usually means gravitational potential energy.
Activity 7.1: a heavy ball dropped into loose sand from about 1 m and then 2 m makes the deepest depression from the greatest height, because raising it higher needed more work. The greater the height above the Earth’s surface, the greater the potential energy.
Deriving U = mgh: take the potential energy of an object of mass m lying on the ground to be zero. To raise it gradually to height h, apply an upward force equal to mg. Work done = force × displacement = mg × h = mgh (Eq. 7.7). By the work-energy theorem this work appears as potential energy:
U = mgh (Eq. 7.8), unit joule.
- U = mgh is valid only near the Earth’s surface. Far from the Earth, g decreases.
- Moving horizontally at constant velocity: U does not change. Raising it: U increases.
- The gain in potential energy depends only on the vertical height gained, whatever path (lift or stairs) is taken.
- Enrichment: work against gravitational, electric or magnetic forces can be stored as potential energy. Work done against friction is not stored.
5. Conservation of Mechanical Energy
Mechanical energy = kinetic energy + potential energy. Follow an object of mass m dropped from rest at point A, height h (NCERT Fig. 7.19).
| Point | Potential energy | Kinetic energy | Mechanical energy |
|---|---|---|---|
| A (height h, t = 0, u = 0) | mgh | 0 | mgh |
| B (after time t, height h′) | mgh′ = mgh − ½mg2t2 | ½mv2 = ½mg2t2 | mgh |
| Just before the ground | 0 | mgh | mgh |
Working for point B: with acceleration g, v = u + gt = gt. Distance fallen: h − h′ = 0 + ½gt2, so h′ = h − ½gt2. Then U = mgh′ = mgh − ½mg2t2 (Eq. 7.10a) and K = ½m(gt)2 = ½mg2t2 (Eq. 7.10b). Their sum is mgh (Eq. 7.10c).
The kinetic energy gained, ½mg2t2, equals the potential energy lost. As an object moves under the gravitational force, its mechanical energy stays the same if no other external force acts on it. This is the conservation of mechanical energy.
The pendulum (Activity 7.2)
Draw a horizontal line on paper behind a simple pendulum. Pull the bob aside to point P on that line and release it.
- At P (extreme): no kinetic energy, only potential energy mgh, where h is the height above the lowest point.
- At Q (lowest point): potential energy zero, only kinetic energy.
- At R (other extreme): kinetic energy zero, potential energy regained.
The bob rises almost to the line, so mechanical energy stays constant. It finally stops because of energy loss due to friction at the support and air resistance.
Why the energy method is powerful
Energy conservation gives final speeds without working through every intermediate step. For a child on a frictionless slide of height h, mgh = ½mv2, so v = √(2gh). The speed at the bottom depends only on the height: the shape of the slide and the mass of the child do not matter. Two children of different masses on the same slide reach the bottom with the same speed, and the tallest slide gives the largest speed.
When friction or air resistance acts, mechanical energy decreases, and the work-energy theorem handles it: work done by the resistive force = change in (K + U). This is how the escape-ramp problem (Example 7.9) is solved.
6. Power
Running up the stairs with your bag in one minute and walking up in five minutes need the same work. The difference is power.
Power is the rate at which work is done. Average power:
P = W / t (Eq. 7.11)
- SI unit: watt (W); 1 W = 1 J s−1, one joule of work done per second. Named after James Watt, inventor of an efficient steam engine.
- More work in the same time needs more power. The same work in less time also needs more power.
- Horsepower (hp): another unit, used for car engines and water pumps. 1 hp = 746 W.
- Do not confuse the symbol W for work with the unit W (watt). Read the context.
Lifting the same mass twice as high in twice the time needs twice the energy but the same power. Raising a flag twice as fast halves the time and doubles the power. Ignoring friction, speed changes the power and leaves the work unchanged. The work to raise a flag with a pulley depends on the flag’s mass, g and the height of the pole.
7. Simple Machines, Mechanical Advantage, Pulley and Inclined Plane
The total work needed for a task cannot be reduced, but the task can be made easier by changing the magnitude or direction of the force applied. Devices that do this are simple machines.
- Effort: the force we apply to a machine.
- Load: the force that needs to be overcome.
- Mechanical advantage = load / effort (Eq. 7.12). It has no unit. MA greater than 1 lets a small effort move a large load.
Pulley
A pulley is a wheel with a groove that guides a rope. With a fixed pulley on a flagpole, you pull down and the flag goes up. Effort equals load, so MA of a fixed pulley = 1. It helps by changing the direction of the effort, since pulling down is easier than lifting up.
Enrichment: in a movable pulley system the load hangs from a moving pulley, one rope end is fixed and effort is applied at the other. Such systems can have MA greater than 1, and are used in elevators and cranes.
Inclined plane
Activity 7.3: a spring balance shows the cart’s full weight when it is lifted straight up to a stool about 0.5 m high, and a smaller reading when it is pulled up a smooth plank. A less steep plank needs even less force, over a larger distance.
Let the load be the weight mg, the effort F′, the length of the incline L and its height h. Moving at constant speed, ignoring friction:
- Work done by you = F′ × L
- Potential energy gained = mgh
- Work-energy theorem: F′ × L = mgh, so mg / F′ = L / h
MA of an inclined plane = load / effort = L / h (Eq. 7.13)
Since L is larger than h, the effort F′ is less than mg and MA is greater than 1. A longer, shallower plane reduces the effort further. Force × displacement stays the same. That is why hill roads wind in gentle slopes and an inclined ladder is easier to climb than a vertical one.
8. The Lever
Activity 7.4: a 30 cm scale resting on a pencil lets light erasers lift a heavier stapler. The scale acts as a lever.
A lever is a rigid bar that can rotate about a fixed point. It has three parts:
- Fulcrum: the fixed point about which the lever rotates.
- Load: the force to be overcome.
- Effort: the force applied.
The distance of the load from the fulcrum is the load arm; the distance of the effort from the fulcrum is the effort arm.
Why a small force can lift a large load
The small force F1 moves a large distance d1; the other end applies a large force F2 through a small distance d2. The work done at one end is transferred to the other:
F1 × d1 = F2 × d2 (Eq. 7.14)
A lever reduces the force needed. It does not reduce the total work done.
Beam balance (Activity 7.5) and the law of the lever
A scale hung at its midpoint (the fulcrum) carries paper-cup pans with identical coins. With more coins in the right pan, that pan must move closer to the centre to balance. The beam balances when n1 × L1 = n2 × L2:
effort × effort arm = load × load arm (Eq. 7.15)
MA of a lever = load / effort = effort arm / load arm (Eq. 7.16)
A longer effort arm means a smaller effort, moving through a larger distance, so the total work stays the same.
Everyday uses: a spoon used to prise open a can lid works as a lever with a long effort arm. To cut something hard with scissors, you push it closer to the fulcrum (the rivet): the load arm becomes shorter, so the same effort produces a larger force on the object.
Classes of levers (enrichment, Table 7.2)
| Class | What lies in between | Examples in NCERT |
|---|---|---|
| Class I | Fulcrum between effort and load | Tongs (scissor-type, pivot in the middle), scissors, crowbar, pliers, balance scale, seesaw |
| Class II | Load between fulcrum and effort | Lemon squeezer, wheel barrow, bottle opener |
| Class III | Effort between fulcrum and load | Tongs (hinged at one end), tweezers, broom, hammer, oar |
NCERT lists tongs under both Class I and Class III. Decide from where the pivot is: at the middle it is Class I, at the end it is Class III.
Many machines are made of two or more simple machines. In all of them, ignoring friction, the work put in equals the useful work done on the load. Machines create no energy. They help us use energy more effectively. So no perpetual motion machine works: real machines lose energy to friction and air resistance and, with no energy input, eventually stop.
Science and society: the Himalayan gharat (panchakki) turns the potential energy of water into kinetic energy in a pipe, which drives a wheel and grinding stone. Dams convert the potential energy of stored water in the same way to generate electricity.
Formula and Reaction Sheet
| Quantity / Law | Formula | Unit |
|---|---|---|
| Work by a constant force | W = F × s | J |
| Work from a graph | W = area under force-displacement graph | J |
| Work-energy theorem | work done on object = change in its energy | J |
| Work by constant force, u to v | W = ½m(v2 − u2) | J |
| Kinetic energy | K = ½mv2 | J |
| Potential energy | U = mgh | J |
| Mechanical energy | E = K + U, constant under gravity alone | J |
| Speed at the bottom of height h | v = √(2gh) | m s−1 |
| Average power | P = W / t | W (watt) = J s−1 |
| Mechanical advantage | MA = load / effort | none |
| Inclined plane | F′ × L = mgh; MA = L / h | none |
| Lever | F1 × d1 = F2 × d2; effort × effort arm = load × load arm | MA: none |
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Important Definitions
- Work done by a constant force: the product of the force applied on an object and the displacement of the object in the direction of the force, W = F × s.
- One joule: the work done on an object when a constant force of 1 newton is applied to it and it is displaced by 1 metre in the direction of the force.
- Energy: the capacity to do work. An object having the capacity to do work is said to possess energy.
- Work-energy theorem: the work done on an object (or a system) is equal to the change in its energy.
- Mechanical energy: the energy an object possesses due to its motion or position; the sum of its kinetic energy and potential energy.
- Kinetic energy: the energy possessed by an object due to its motion, K = ½mv2.
- Potential energy: the energy stored by an object as a result of its deformation, or in a system of objects due to their relative positions.
- Conservation of mechanical energy: the mechanical energy of an object moving under the gravitational force stays constant if no other external force acts on it.
- Power: the rate at which work is done, P = W/t.
- Simple machine: a device that makes a task easier by changing the magnitude or direction of the force that needs to be applied, without reducing the total work.
- Effort: the force applied to a machine. Load: the force that needs to be overcome.
- Mechanical advantage: the ratio of the load to the effort.
- Lever: a rigid bar that can rotate about a fixed point called the fulcrum.
- Load arm / effort arm: the distance of the load / effort from the fulcrum.
Solved Examples (NCERT-Based)
Example 1: Negative work by a goalkeeper (NCERT Example 7.2)
Question: While saving a goal, a goalkeeper’s hand moved back by 15 cm as she stopped a ball while applying a force of 200 N. How much work did the goalkeeper do on the ball in stopping it?
Solution: The force is opposite to the ball’s motion, so the displacement along the force is negative: s = −15 cm = −0.15 m.
W = F × s = 200 N × (−0.15 m) = −30 J. The goalkeeper does 30 J of negative work, which removes 30 J of kinetic energy from the ball.
Example 2: Kinetic energy of a fast delivery (NCERT Example 7.5)
Question: In one of their fastest deliveries, an Indian cricketer bowled a cricket ball with an approximate mass of 0.2 kg at a velocity of about 154.8 km h−1. Calculate the kinetic energy of the ball at the time of its delivery.
Solution: v = 154.8 ÷ 3.6 = 43 m s−1.
K = ½mv2 = ½ × 0.2 kg × (43 m s−1)2 = 0.1 × 1849 = 184.9 J.
Example 3: Stopping a jet on an aircraft carrier (NCERT Example 7.6)
Question: A jet aircraft of mass 15000 kg lands on the deck of an aircraft carrier. To stop the aircraft within the short length of the deck a hook on the aircraft’s tail is caught in a wire stretched across the deck. The wire exerts an approximately constant backward force of 367500 N and stops the jet within 100 m. What was the velocity of the aircraft just before the wire caught the hook?
Solution: Let the speed be v.
- Initial K = ½ × 15000 × v2; final K = 0. Change in K = −½ × 15000 × v2.
- The wire’s force is opposite to the displacement, so its work is negative: W = 367500 N × (−100 m) = −36750000 J.
- Work-energy theorem: −½ × 15000 × v2 = −36750000, so 7500 v2 = 36750000.
- v2 = 4900 m2 s−2, v = 70 m s−1 = 70 × 3.6 = 252 km h−1.
Example 4: Truck on an escape ramp (NCERT Example 7.9)
Question: Escape ramps are inclined planes filled with sand or gravel that help stop trucks when their brakes fail on a highway. A truck of mass 10000 kg is moving at 72 km h−1 when its brakes fail. The driver steers it onto an escape ramp inclined at 30°, where the truck comes to a rest. If the sand exerts a force of 50000 N opposite to truck’s motion, what is the minimum length of the ramp to be able to stop such a truck? Take g = 10 m s−2 (Hint: For a 30° incline, the truck rises 1 m vertically for every 2 m it travels along the ramp).
Solution:
- v = 72 km h−1 = 20 m s−1. Initial K = ½ × 10000 × 202 = 2000000 J; initial U = 0. Initial energy = 2000000 J.
- Let the truck travel d along the ramp. Height gained = d/2. Final K = 0; final U = 10000 × 10 × d/2 = 50000 N × d.
- Change in energy = 50000d − 2000000.
- Work by the sand = −50000 N × d (force opposite to motion).
- Work-energy theorem: −50000d = 50000d − 2000000, so 100000d = 2000000, d = 20 m.
Example 5: Power of a weightlifter (NCERT Example 7.10)
Question: A weightlifter lifts a 75 kg mass by 2 m in 5 seconds. How much power would she require for this task?
Solution: W = mgh = 75 kg × 10 m s−2 × 2 m = 1500 J. P = W/t = 1500 J / 5 s = 300 W.
Example 6: Power of a car engine (NCERT Example 7.11)
Question: A car of mass 1000 kg starts from rest and reaches a speed of 72 km h−1 in 10 seconds. Calculate the power of the engine required to achieve this start.
Solution: v = 72000 m / 3600 s = 20 m s−1, u = 0.
Work done by the engine = change in K = ½ × 1000 × 202 − 0 = 200000 J.
P = 200000 J / 10 s = 20000 W (20 kW).
Example 7: Mechanical advantage of a ramp (NCERT Example 7.12)
Question: A person uses an inclined ramp to raise an object over a step 30 cm high. The ramp has a width of 40 cm. What is the mechanical advantage of the ramp that helps the person achieve the task?
Solution: Height (30 cm) and horizontal width (40 cm) are the two perpendicular sides of a right-angled triangle; the ramp is the hypotenuse. L = √(302 + 402) = √2500 = 50 cm.
MA = L / h = 50 cm / 30 cm = 1.67.
Example 8: Balancing a seesaw (NCERT Example 7.13)
Question: For a seesaw having four seats A, B, D, E and fulcrum at C, AC = EC = 2 m and BC = DC = 1 m. On which seats should children of masses 15 kg and 30 kg sit to make the seesaw balanced?
Solution: A and B lie on one side of the fulcrum C, D and E on the other. Put the 15 kg child on A (2 m from C) and the other child at distance L on the other side. Using effort × effort arm = load × load arm: 15 kg × 2 m = 30 kg × L, so L = 1 m. The 30 kg child sits on seat D. (Equally, 15 kg on E and 30 kg on B.) The heavier child sits closer to the fulcrum.
Example 9: Ball thrown up against air resistance (NCERT exercise 10)
Question: 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).
Solution: (i) Gravity acts downward. Going up, displacement is upward: negative work. Coming down, displacement is downward: positive work.
(ii) Initial energy: K = ½ × 2 × 202 = 400 J, U = 0. At the top: K = 0, U = 2 × 10 × 19.4 = 388 J. Change in mechanical energy = 388 − 400 = −12 J. Work done by air resistance = −12 J.
Competency-Based Questions (with answers)
1. Source-based: The gharat of the hills
In the Himalayan region, water flowing downhill converts its potential energy into kinetic energy. The traditional gharat or panchakki uses this: water comes down a pipe, drives a wheel, and the wheel turns a grinding stone for grain.
(a) Name the energy changes from the top of the pipe to the stone. Potential energy of water → kinetic energy of water → kinetic (rotational) energy of the wheel and stone, with some heat and sound due to friction.
(b) 50 kg of water falls through a vertical height of 8 m. How much potential energy does it lose? (g = 10 m s−2) mgh = 50 × 10 × 8 = 4000 J.
(c) With no losses, what would its speed be at the bottom? v = √(2gh) = √160 ≈ 12.6 m s−1.
2. Data-based: Beam balance readings
In Activity 7.5, the left pan holds 1 coin at 20 cm from the fulcrum. A student balances the beam with the right pan as follows.
| Coins in right pan, n2 | 1 | 2 | 4 | 8 |
|---|---|---|---|---|
| Distance of right pan, L2 (cm) | 20 | 10 | 5 | ? |
(a) What rule do the readings follow? n1 × L1 = n2 × L2 = 20 coin cm in every case.
(b) Find the missing value. 8 × L2 = 20, L2 = 2.5 cm.
(c) Treating the single coin as effort, what is the MA in the last case? MA = load/effort = 8/1 = 8 = effort arm/load arm = 20/2.5.
3. Assertion-Reason
Assertion (A): A girl walking on a level road with a box in her hands does no work on the box through the upward force she applies. Reason (R): Work done by a force is zero when the force is perpendicular to the displacement.
Answer: (a) Both true, and R is the correct explanation of A. Her force is vertical; the box moves horizontally.
4. Assertion-Reason
Assertion (A): When the speed of a car doubles, its kinetic energy doubles. Reason (R): Kinetic energy is proportional to the square of the speed.
Answer: (d) A is false, R is true. Doubling the speed makes K four times.
5. Assertion-Reason
Assertion (A): A fixed pulley is useful even though its mechanical advantage is 1. Reason (R): A fixed pulley changes the direction of the effort, and pulling down is easier than lifting up.
Answer: (a) Both true, and R is the correct explanation of A.
6. Assertion-Reason
Assertion (A): On a frictionless slide, a heavy child and a light child reach the bottom with the same speed. Reason (R): Both the potential energy at the top and the kinetic energy at the bottom are proportional to the mass.
Answer: (a) Both true, and R is the correct explanation of A. m cancels in mgh = ½mv2, leaving v = √(2gh).
7. Case-based: Roads on hills
A mountain road climbs 100 m of height over 2 km of road length. A 10000 kg truck moves up it at a steady speed. Ignore friction; g = 10 m s−2.
(a) Treat the road as an inclined plane. What is its MA? MA = L/h = 2000/100 = 20.
(b) What force along the road is needed? Effort = load/MA = (10000 × 10)/20 = 5000 N.
(c) Compare the work done with lifting the truck straight up 100 m. Along the road: 5000 N × 2000 m = 107 J. Straight up: mgh = 100000 × 100 = 107 J. The same work; the gentle slope only lowers the force needed.
8. Case-based: Two students on the stairs
Riya (40 kg) runs up a 6 m high staircase in 12 s. Aman (60 kg) walks up the same stairs in 30 s. Take g = 10 m s−2.
(a) Who does more work? Riya: 40 × 10 × 6 = 2400 J. Aman: 60 × 10 × 6 = 3600 J. Aman does more work.
(b) Who develops more power? Riya: 2400/12 = 200 W. Aman: 3600/30 = 120 W. Riya is more powerful.
(c) What does this show? Work depends on force and distance; power also depends on how quickly the work is done.
Important Questions for Exams
1-Mark Questions
- State whether True or False: Work is said to be done when a force is applied, even if the object does not move. False. With no displacement, W = 0.
- Fill in the blank: 1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force. 1 (one).
- State whether True or False: A motionless stretched rubber band has kinetic energy. False. It has potential energy due to its deformed shape.
- What is the mechanical advantage of a fixed pulley? 1, since effort equals load.
- Express 1 horsepower in watt. 1 hp = 746 W.
2-Mark Questions
- 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. (iii) and (iv). Gravity still acts, so force and acceleration (g, downward) are not zero; the ball is momentarily at rest at its greatest height.
- Two objects A and B of mass m and 4m have the same kinetic energy. What is the ratio of the magnitude of velocities of A and B? ½mvA2 = ½(4m)vB2 gives vA = 2vB, so 2 : 1.
- Explain why roads on hills are built to wind around in gentle slopes rather than going straight up. A winding road is an inclined plane with a larger length L for the same height h. MA = L/h is larger, so vehicles need a smaller force, though they travel a longer distance.
3-Mark Questions
- 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? (i) and (ii) both: mgh = 50 × 10 × 72.5 = 36250 J. (iii) It depends only on the height gained, not on the path.
- 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. Assume both lifts start from the ground and floor n is n floor-heights up, so the 20th floor is twice as high as the 10th. First lift: energy E = mgh in time t, power E/t. Second lift: energy mg(2h) = 2E, so twice the energy (E more). Power = 2E/2t = E/t: the same power, no extra power. (If the second lift started from the 10th floor, it would need energy E in time 2t, so half the power.)
- 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. Fuel ∝ kinetic energy given ∝ total mass at the same v. Day 1: 160 kg; day 2: 200 kg. Ratio = 160 : 200 = 4 : 5.
5-Mark Questions
- Derive K = ½mv2 using the work-energy theorem. How does K change if the velocity is doubled? A constant force F acts on mass m, changing velocity from u to v over displacement s. From v2 = u2 + 2as, s = (v2 − u2)/(2a). W = F × s = ma × (v2 − u2)/(2a) = ½m(v2 − u2). By the work-energy theorem this is the change in energy. Starting from rest (u = 0) it is the kinetic energy: K = ½mv2. Doubling v makes K = ½m(2v)2 = 4 × ½mv2, four times the original.
- Show that the mechanical energy of a freely falling object remains constant. Drop mass m from rest at height h. At the top: U = mgh, K = 0. After time t: v = gt and height h′ = h − ½gt2, so U = mgh − ½mg2t2 and K = ½mg2t2. Just before the ground: v2 = 2gh, so K = mgh and U = 0. At every point K + U = mgh: the potential energy lost equals the kinetic energy gained.
- 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. (i) v = √(2gh) = √(2 × 10 × 10) = √200 = 10√2 ≈ 14.1 m s−1. (ii) Kinetic energy on arrival = mgh = 1.5 × 10 × 10 = 150 J. The question says all of it is used to make the depression, so the work done against the sand’s 3000 N force equals 150 J: 3000 N × depth = 150 J, depth = 0.05 m = 5 cm. This treats the sand force as the only force while sinking; the tiny extra potential energy lost in the 5 cm (15 N × 0.05 m ≈ 0.75 J) is ignored.
Common Mistakes and Examiner Tips
- Calling any effort “work”. Holding a heavy bag still or pushing a wall does zero work on it. Fix: check for displacement along the force before writing W.
- Using total distance when the force is perpendicular. Carrying a box across a level floor: the upward force does no work. Fix: use only the displacement in the direction of the force.
- Dropping the minus sign. Braking, friction, a goalkeeper stopping a ball, gravity on a rising ball: all do negative work. Fix: take s negative when it is opposite to the force, and keep the sign through the work-energy equation.
- Squaring km h−1. Fix: convert to m s−1 first (divide by 3.6) and only then use ½mv2.
- Assuming K doubles when v doubles. Fix: K ∝ v2, so double speed gives four times K.
- Grams in mgh. Fix: 200 g is 0.2 kg; convert to SI before substituting.
- Thinking the stairs give more potential energy than the lift. Fix: U = mgh depends only on the height gained, never on the path.
- Saying force or acceleration is zero at the top of a throw. Only the velocity (and so K) is zero; gravity still acts with acceleration g. Fix: separate velocity from acceleration.
- Believing a machine reduces work. Fix: a machine reduces or redirects the force; the effort moves further, and the total work (ignoring friction) stays the same.
- Inverting mechanical advantage. Fix: MA = load/effort = L/h for a ramp = effort arm/load arm for a lever.
Quick Revision Points
- Work by a constant force = force × displacement in the direction of the force: W = F × s.
- SI unit of work: joule; 1 J = 1 N × 1 m = 1 kg m2 s−2.
- Work has no direction; it is positive, negative or zero.
- W = 0 when F = 0, when s = 0, or when the force is perpendicular to the displacement.
- Work-energy theorem: work done on an object = change in its energy.
- K = ½mv2; derived from W = ½m(v2 − u2) with u = 0.
- Double speed: four times K. Same K, mass 4 times: speed half.
- Potential energy is stored by deformation (spring, bow, slingshot) or by relative position (Earth-ball, magnets, charges).
- U = mgh near the Earth’s surface; it depends on height, not path.
- Only gravity acting: K + U stays constant. Falling object: U lost = K gained.
- Frictionless slide or free fall through h: v = √(2gh), independent of mass and shape of path.
- Power = rate of doing work, P = W/t; 1 W = 1 J s−1; 1 hp = 746 W.
- Simple machines change the magnitude or direction of force; they never reduce total work.
- MA = load / effort. Fixed pulley: MA = 1, changes direction only.
- Inclined plane: F′ × L = mgh; MA = L/h > 1; longer and gentler means less effort.
- Lever: effort × effort arm = load × load arm; MA = effort arm / load arm.
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams (periodic tests, half-yearly and annual papers), and the NCERT source gives no fixed mark allocation for it.
| Question type | What this chapter typically supplies |
|---|---|
| MCQ, true/false, fill in the blanks | Units (joule, watt, horsepower), sign of work, K and U formulas, MA of a fixed pulley |
| Assertion-Reason | Zero work for perpendicular force, K ∝ v2, mass-independence of v = √(2gh), machines and total work |
| Short answer (2 to 3 marks) | One-step K, U, power and MA numericals; energy transformations; path-independence of mgh |
| Long answer (5 marks) | Derivation of K = ½mv2, proof of conservation of mechanical energy, work-energy problems with friction or air resistance |
| Case-based and source-based | Escape ramps, water mills, hill roads, seesaws and beam balances, stair-climbing power |
This chapter is the foundation for Class 10 physics: the work-energy theorem and energy conservation return in every later numerical chapter.
Class 9 Science · Chapter 7 – swipe through all 10 cards to understand the whole chapter.
Work Done by a Force
Work is force times the displacement in the direction of the force.
Name the force doing work and the object it is done on.
- Lifting 3 bags together or 1 bag 3 times as high: 3 times the work.
- 1 J = 1 kg m2 s⁻2; work has no direction, only a sign.
- Area under a force-displacement graph gives the work done.
Positive, Negative, Zero
The sign depends on the directions of force and displacement.
Zero work: no force, no displacement, or force perpendicular to motion.
- Pushing a wheelchair forward: positive work.
- Stopping a ball or braking a car: negative work.
- Pushing a wall or carrying a box on level ground: zero work.
Work-Energy Theorem
Work done on an object equals the change in its energy.
Holds for systems of objects and for varying forces too.
- Energy is the capacity to do work, measured in joule.
- Positive work adds energy; negative work removes it.
- Jet stopped by a 367500 N wire in 100 m was landing at 70 m/s.
Kinetic Energy
Energy an object has because it is moving.
Convert km/h to m/s before squaring.
- Double the speed: four times the kinetic energy.
- 0.2 kg ball at 43 m/s has K = 184.9 J.
- Same K with 4 times the mass means half the speed.
Potential Energy
Energy stored by deformation or by the relative position of objects.
Depends on height gained, not on the path taken.
- Stretched slingshot, bent bow and compressed spring store energy.
- Earth-ball, magnets and charges store energy by separation.
- 0.2 kg ball at 10 m: U = 0.2 × 10 × 10 = 20 J.
Mechanical Energy Is Conserved
Under gravity alone, kinetic plus potential energy stays constant.
Friction and air resistance remove mechanical energy as heat.
- Falling object: U lost at every point equals K gained.
- Pendulum: only U at the extremes, only K at the lowest point.
- Slide speed depends only on height, not on mass or shape.
Power
Power is the rate at which work is done.
Same work in less time needs more power.
- Lifting 75 kg by 2 m in 5 s needs 1500 J / 5 s = 300 W.
- Car of 1000 kg to 20 m/s in 10 s needs 20000 W.
- Double the height in double the time: same power.
Mechanical Advantage and Pulley
Simple machines change the size or direction of a force.
Total work is never reduced by a machine.
- Effort is the force applied; load is the force overcome.
- Fixed pulley: MA = 1, it only changes the direction of effort.
- Movable pulleys can give MA greater than 1.
Inclined Plane
A slope lets a smaller force raise a load through a longer distance.
A longer, gentler slope needs less effort.
- Ramp 50 cm long over a 30 cm step: MA = 50/30 = 1.67.
- Hill roads wind in gentle slopes to reduce the force needed.
- An inclined ladder is easier to climb than a vertical one.
The Lever
A rigid bar that turns about a fixed point called the fulcrum.
MA = effort arm / load arm.
- Seesaw: 15 kg at 2 m balances 30 kg at 1 m.
- Class I fulcrum, Class II load, Class III effort in between.
- Scissors cut hard things near the rivet: shorter load arm.
📝 Practice Work, Energy, and Simple Machines - 10 questions
Concept-check questions · with answers & explanations
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Why A: Work counts only the displacement in the direction of the force. The girl’s force on the box is upward (balancing its weight) while the box moves horizontally, so the displacement along the force is zero and W = F × 0 = 0. The reason states exactly this cause, so it correctly explains the assertion.
Why not B: The perpendicular directions are precisely why the work is zero, so the reason is the explanation, not an unrelated fact.
Why not C: The reason is a correct statement from the chapter: perpendicular force and displacement give zero work.
Why not D: The assertion is true; the girl may feel tired, but scientifically her upward force does no work on the box.
Remember: Only displacement along the force counts; sideways motion under an upward force is zero work.
Why A: Step 1 (equate kinetic energies, K = ½ m v²): (1)/(2) m v_A² = (1)/(2) (4m) v_B², so v_A² = 4 v_B². Step 2 (take the square root): v_A = 2 v_B, so v_A : v_B = 2 : 1. The lighter object must move faster to carry the same energy.
Why not B: 4 : 1 is the ratio of the squares of the velocities, v_A² : v_B²; the square root was not taken.
Why not C: 1 : 4 simply copies the mass ratio m : 4m; for equal kinetic energy the velocity ratio is the inverse square root of the mass ratio.
Why not D: 1 : 2 would make the lighter object slower; with equal energy the lighter one must be faster, not slower.
Remember: For equal K, v is proportional to 1/√m, so four times the mass means half the speed.
Why B: U = mgh = 50 kg × 10 m s⁻² × 72.5 m = 36250 J. Only the mass, g and the vertical height h appear in the formula; the route (elevator or zig-zag stairs) does not. So the gain in potential energy is 36250 J both times.
Why not A: Whoever does the work, the student ends up 72.5 m higher and so has gained 36250 J of potential energy.
Why not C: A longer path means more distance walked, but potential energy counts only the vertical height gained, not the length of the path.
Why not D: Being lifted directly changes who does the work, not how much potential energy is stored for a given height.
Remember: Potential energy gain = mgh, whatever the route to that height.
Why A: Step 1 (potential energy at the top, U = mgh): U = 1.5 × 10 × 10 = 150 J, which becomes kinetic energy just before impact. Step 2 (work-energy theorem for the sand): the sand’s resistive force must remove all 150 J, so 3000 N × d = 150 J. Step 3 (solve and convert): d = 150 / 3000 = 0.05 m = 5 cm.
Why not B: 50 cm is ten times too large, from 1500/3000 instead of 150/3000.
Why not C: 0.5 cm converts 0.05 m wrongly; 0.05 m is 5 cm because 1 m = 100 cm.
Why not D: 2 m comes from dividing 3000 by 1500, inverting the ratio of energy to force.
Remember: Energy lost = resistive force × distance moved against it.
Why C: The work done on the bag is its weight × the height of the stairs, which is identical on both days. Power = work / time, so doing that work in 1 minute instead of 5 minutes needs 5 times the power. This is the example the chapter uses to introduce power.
Why not A: Taking longer for the same work means a lower rate of doing work, so less power, not more.
Why not B: Same work in a shorter time is a higher rate, so the power is not the same.
Why not D: Speed of climbing does not change the force or the height, so the work is the same and the bag ends up with the same potential energy.
Remember: Same work, less time, more power.
Why A: The chapter states that a lever reduces the force required for a task but not the total work done. The end where the small effort F1 acts moves a large distance d1 while the load end moves a small distance d2, with F1 × d1 = F2 × d2. The reason gives precisely this mechanism, so it correctly explains the assertion.
Why not B: The distance trade-off is the very cause of the work staying constant, so the reason is the explanation, not a separate fact.
Why not C: The reason is correct; the work done on one end of the lever is transferred to the other end, F1 × d1 = F2 × d2.
Why not D: The assertion is correct; machines do not create energy, they only help us use it more effectively.
Remember: Machines trade force for distance; the work stays the same.
Why D: Step 1 (unit conversion): 15 cm = 0.15 m. Step 2 (sign of work): the ball keeps moving forward while her force points backward, so the displacement in the direction of the force is -0.15 m. Step 3 (W = F × s): W = 200 N × (-0.15 m) = -30 J.
Why not A: +3000 J uses 15 instead of 0.15, treating centimetres as metres, and also drops the negative sign.
Why not B: -3000 J has the correct sign but multiplies 200 N by 15 rather than by 0.15 m.
Why not C: +30 J has the right magnitude but the wrong sign; a force opposite to the displacement does negative work on the object.
Remember: Force against the motion means negative work; convert cm to m before multiplying.
Why C: Step 1 (initial kinetic energy, K = ½ m v²): K = ½ × 1000 × 20² = 500 × 400 = 200000 J. Step 2 (work-energy theorem): work done by the brakes = change in kinetic energy = 0 – 200000 J = -200000 J. The brakes act against the motion, so the work is negative and the kinetic energy of the car is removed.
Why not A: -10000 J is ½ × 1000 × 20, using the velocity unsquared.
Why not B: +200000 J has the right size but the wrong sign; a force opposite to the displacement does negative work on the car.
Why not D: -400000 J uses m v² without the ½.
Remember: A stopping force does negative work equal in size to the kinetic energy it removes.
Why D: U = mgh depends on the height h above the ground. Moving horizontally at constant velocity keeps h fixed, so the potential energy does not change. Raising the ball vertically increases h, so mgh increases, because work is done against gravity and stored (Pause and Ponder 6).
Why not A: Horizontal motion does not change the height, so no potential energy is gained there; the velocity being constant also means no change in kinetic energy.
Why not B: Raising the ball vertically does work against gravity, and by the work-energy theorem that work is stored as potential energy.
Why not C: This swaps the two cases; height changes in the vertical raise, not in the horizontal motion.
Remember: Same height, same potential energy; higher up, more potential energy.
Why C: Step 1 (total mechanical energy at release, U = mgh): E = 2 × 10 × 20 = 400 J, with K = 0. Step 2 (potential energy at 5 m): U = 2 × 10 × 5 = 100 J. Step 3 (conservation of mechanical energy): K = E – U = 400 – 100 = 300 J.
Why not A: 200 J assumes the energy is split equally between kinetic and potential, which happens only at the halfway height of 10 m.
Why not B: 100 J is the potential energy remaining at 5 m, not the kinetic energy.
Why not D: 400 J is the total mechanical energy; the object still has 100 J of potential energy at 5 m, so not all of it is kinetic yet.
Remember: Kinetic energy at any height = mg × (height fallen so far).
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Frequently Asked Questions
Work done W = F × s is zero in three cases: when no force acts (F = 0), when the object does not move (s = 0), as when you push a rigid wall or hold a barbell steady, and when the force is perpendicular to the displacement, as when you carry a box while walking on level ground. You may feel tired, but no work is done on the object.
Work is positive when the displacement is in the same direction as the force, such as pushing a wheelchair forward. It is negative when the displacement is opposite to the force, such as a goalkeeper stopping a ball or brakes stopping a car. Positive work increases the object’s energy; negative work decreases it.
The work done on an object equals the change in its energy. It also holds for a system of objects and for forces that are not constant. Using it with a constant force and v² = u² + 2as gives W = ½m(v² − u²), which leads to the kinetic energy formula K = ½mv² for an object starting from rest.
Kinetic energy K = ½mv² depends on the square of the speed. If the speed doubles, K becomes ½m(2v)² = 4 × ½mv², so it becomes four times the original value. If the mass doubles at the same speed, K only doubles. Always convert km/h to m/s before squaring the speed.
No. Near the Earth’s surface U = mgh depends only on the mass and the height gained. A 50 kg student reaching the top of a 72.5 m building by lift or by stairs gains the same 50 × 10 × 72.5 = 36250 J of potential energy. Only the vertical height matters.
Mechanical energy is the sum of kinetic and potential energy. For an object moving under gravity alone, this sum stays constant: potential energy lost equals kinetic energy gained. A falling object has mgh at the top, and at every point below K + U is still mgh. Friction and air resistance reduce mechanical energy, which is why a pendulum finally stops.
Power is the rate at which work is done, P = W/t. Its SI unit is the watt, where 1 W = 1 J s⁻¹. Doing the same work in less time needs more power. Another unit, the horsepower, is used for car engines and water pumps, and 1 hp = 746 W.
No. Simple machines make a task easier by changing the magnitude or direction of the force, but the total work stays the same when friction is ignored. With a ramp or lever the effort is smaller but it moves through a larger distance. Mechanical advantage = load/effort; for a ramp it is L/h, for a lever effort arm/load arm.
