Work and Energy

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CLASS IX Science ~6–8 marks/year Ch 11 of 15
Work and Energy

Class 9 · Science · NCERT chapter notes · Akanksha Classes

Snapshot
  • Work done by a constant force: W = Fs (force parallel to displacement) or W = Fs cosθ (general case).
  • Kinetic energy: KE = ½mv² — derived from Newton’s second law and kinematics.
  • Potential energy (gravitational): PE = mgh — energy stored by virtue of position above ground.
  • Work–energy theorem: net work done on an object = change in its kinetic energy, W = ΔKE.
  • Law of conservation of energy: total mechanical energy (KE + PE) is constant in the absence of non-conservative forces; energy can only be converted, never created or destroyed.
  • Power: P = W/t = Fv; unit is Watt (W); 1 hp = 746 W.
  • Commercial unit: 1 kilowatt-hour (kWh) = 3.6 × 10&sup6; J (also called 1 “unit” of electricity).
  • Board weightage: ~6–8 marks/year — typically one numerical (2–3 marks), one derivation (3 marks), one MCQ or definition (1 mark).
Detailed notes

1. Scientific definition of work

In everyday language we say a person “works” even when standing still holding a heavy bag. In physics, work is done only when a force causes a displacement in the body on which it acts.

Definition: Work done by a force F on an object that undergoes a displacement s in the direction of the force is:

$$W = F \times s$$

When the force and displacement are not in the same direction, the component of force along the displacement does the work. If θ is the angle between the force vector and the displacement vector:

$$W = F\,s\cos\theta$$
  • θ = 0° (force and displacement same direction): W = Fs (maximum positive work).
  • θ = 90° (force perpendicular to displacement): W = 0 (no work done — e.g., a person carrying a bag horizontally while walking, or the Moon orbiting Earth).
  • θ = 180° (force and displacement opposite): W = −Fs (negative work — e.g., friction opposing motion, brakes stopping a car).

Three conditions for work to be done:

  1. A force must be applied.
  2. The object must be displaced.
  3. There must be a component of force in the direction of displacement (i.e., cosθ ≠ 0).

Positive work is done when force and displacement are in the same direction (e.g., pushing a block forward). Negative work is done when force and displacement are opposite (e.g., friction, gravity when ball is thrown upward). Zero work — even though force and displacement both exist, if they are perpendicular, no work is done (e.g., gravity on a horizontally moving object).

2. Unit of work — the Joule

The SI unit of work is the Joule (J), named after James Prescott Joule.

$$1 \text{ J} = 1 \text{ N} \times 1 \text{ m} = 1 \text{ kg m}^2 \text{ s}^{-2}$$

1 Joule is the work done when a force of 1 Newton displaces an object by 1 metre in the direction of the force.

Larger units: 1 kJ = 1000 J; 1 MJ = 10&sup6; J.

Note: Work is a scalar quantity (it has magnitude only, no direction), even though it is calculated from two vector quantities (force and displacement).

3. Energy — the capacity to do work

Energy is defined as the capacity (or ability) of an object to do work. An object has energy if it can exert a force on another object and cause displacement.

  • Energy is a scalar quantity.
  • SI unit: Joule (J) — same as work.
  • Energy and work are interconvertible; when work is done on a system, energy is stored in it.

Forms of mechanical energy:

  1. Kinetic energy (KE) — energy due to motion.
  2. Potential energy (PE) — energy due to position or configuration.

Total mechanical energy E = KE + PE. In an isolated system (no friction or air resistance), E is constant.

4. Kinetic energy and its derivation

An object of mass m moving with velocity v has kinetic energy:

$$KE = \frac{1}{2}mv^2$$

Derivation (from Newton’s second law and equations of motion):

Consider an object of mass m initially at rest (u = 0) on a frictionless surface. A constant force F acts on it and it moves a distance s, reaching velocity v.

Step 1 — From Newton’s second law:

$$F = ma$$

Step 2 — Work done by the force:

$$W = F \times s = mas$$

Step 3 — From the kinematic equation v² = u² + 2as, with u = 0:

$$v^2 = 2as \implies as = \frac{v^2}{2}$$

Step 4 — Substitute:

$$W = m \times \frac{v^2}{2} = \frac{1}{2}mv^2$$

This work is stored as the kinetic energy of the object: KE = ½mv².

Key points:

  • KE is always positive (mass and v² are both positive).
  • KE depends on speed squared — doubling speed quadruples KE.
  • KE depends on mass — a heavier object moving at the same speed has more KE.
  • When a body decelerates, its KE decreases; when it accelerates, KE increases.

5. Gravitational potential energy

When an object of mass m is raised to a height h above the ground, work is done against gravity. This work is stored as gravitational potential energy:

$$PE = mgh$$

Derivation: To lift the object at constant speed (no acceleration), the applied force must equal the weight: F = mg (upward). The work done over height h:

$$W = F \times h = mgh$$

This equals the potential energy stored.

Key points:

  • PE is measured from a reference level (usually ground). PE = 0 at ground level.
  • g = 9.8 m/s² (use 10 m/s² unless told otherwise in CBSE problems).
  • PE depends on height above the reference — raising an object higher increases PE.
  • When an object falls, PE decreases and KE increases.
  • Elastic potential energy is also stored in a stretched or compressed spring.

6. Work–energy theorem

The work–energy theorem states that the net work done on an object equals the change in its kinetic energy:

$$W_{net} = \Delta KE = KE_f - KE_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$$

Proof sketch: Using F = ma and v² = u² + 2as:

$$W = Fs = mas = m \cdot \frac{v^2-u^2}{2} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$$

Implications:

  • If net work is positive, kinetic energy increases (object speeds up).
  • If net work is zero, kinetic energy remains unchanged (constant speed).
  • If net work is negative, kinetic energy decreases (object slows down).

This theorem connects force (a vector) to energy (a scalar), making many problems far easier to solve.

7. Law of conservation of energy

The Law of Conservation of Energy states: Energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy of an isolated system remains constant.

Mechanical energy conservation: In the absence of friction and air resistance:

$$KE + PE = \text{constant} \implies \frac{1}{2}mv^2 + mgh = \text{constant}$$

Classic example — a ball falling freely from height H:

Position Height PE KE Total E
Top (released from rest) H mgH 0 mgH
Mid-point H/2 mgH/2 mgH/2 mgH
Just before hitting ground 0 0 mgH mgH

At every point, KE + PE = mgH = constant. PE is fully converted to KE by the time the ball reaches the ground.

Pendulum: At the extreme positions (maximum height), all energy is PE; at the mean (lowest) position, all energy is KE. Energy alternates between PE and KE every half-swing.

When friction is present: Some mechanical energy converts to heat (thermal energy). Total energy (mechanical + thermal) is still conserved — but mechanical energy alone decreases. That is why on Earth, a swinging pendulum eventually stops.

Other energy conversions: Electrical energy → light + heat (bulb); chemical energy → kinetic + heat (engine); solar energy → electrical (solar cell). In every case the total energy bookkeeping always balances.

8. Power

Power is the rate at which work is done (or energy is transferred).

$$P = \frac{W}{t} = \frac{\text{Work done}}{\text{Time taken}}$$

Since W = Fs and s/t = v (speed), we also have:

$$P = \frac{Fs}{t} = F \times v$$

This form P = Fv is very useful when force and speed are given directly.

SI unit of power: Watt (W), named after James Watt.

$$1 \text{ W} = 1 \text{ J s}^{-1} = 1 \text{ kg m}^2 \text{ s}^{-3}$$

1 Watt is the power when 1 Joule of work is done in 1 second.

Other units of power:

  • 1 kilowatt (kW) = 1000 W
  • 1 megawatt (MW) = 10&sup6; W
  • 1 horsepower (hp) = 746 W (used for engines)

Power is also a scalar quantity.

9. Average power

When the rate of doing work varies over time, we use average power:

$$P_{avg} = \frac{\text{Total work done}}{\text{Total time taken}} = \frac{W_{total}}{t_{total}}$$

Example: A person does 6000 J of work in 2 minutes.

$$P_{avg} = \frac{6000}{120} = 50 \text{ W}$$

Difference between power and energy:

  • Energy tells us how much work can be done.
  • Power tells us how fast the work is done.
  • A 100 W bulb uses more energy per second than a 60 W bulb, but both can do the same total work given enough time.

10. Commercial unit of energy — the kilowatt-hour

The Joule is too small a unit for measuring household electrical energy consumption. The commercial (practical) unit used by electricity boards is the kilowatt-hour (kWh).

Definition: 1 kWh is the energy consumed by a device of power 1 kW running for 1 hour.

$$1 \text{ kWh} = 1 \text{ kW} \times 1 \text{ h} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$$

On your electricity bill, 1 kWh is called 1 unit. The meter in your home measures energy consumption in kWh.

Example: A 100 W bulb left on for 10 hours:

$$E = P \times t = 100 \text{ W} \times 10 \text{ h} = 1000 \text{ Wh} = 1 \text{ kWh} = 1 \text{ unit}$$

Conversion summary:

  • 1 kWh = 3.6 × 10&sup6; J = 3600 kJ
  • To find units consumed: Units = (Power in W × Time in h) / 1000

11. NCERT Worked Examples

NCERT Example 11.1 — Work done pushing a body

Problem: A force of 5 N is applied on an object and it moves 2 m in the direction of the force. What is the work done?

Given: F = 5 N, s = 2 m, θ = 0°

Solution: W = Fs cosθ = 5 × 2 × cos 0° = 5 × 2 × 1 = 10 J

NCERT Example 11.2 — Zero work done by gravity on horizontal motion

Problem: A man is carrying a bag (10 kg) and walks 5 m on a level road. What is the work done against gravity?

Given: m = 10 kg, displacement s = 5 m (horizontal), gravity acts downward (θ = 90° between gravity and displacement).

Solution: W = mgs cos 90° = 10 × 10 × 5 × 0 = 0 J

No work is done against gravity when displacement is horizontal, even though the person exerts effort to hold the bag.

NCERT Example 11.3 — Finding KE of a moving object

Problem: An object of mass 15 kg is moving with a uniform velocity of 4 m/s. What is its kinetic energy?

Given: m = 15 kg, v = 4 m/s

Solution: KE = ½mv² = ½ × 15 × (4)² = ½ × 15 × 16 = 120 J

NCERT Example 11.4 — Velocity from kinetic energy

Problem: What is the speed of an object of mass 10 kg if its KE is 500 J?

Given: m = 10 kg, KE = 500 J

Solution: KE = ½mv² ⇒ v² = 2KE/m = (2 × 500)/10 = 100 ⇒ v = √100 = 10 m/s

NCERT Example 11.5 — PE of a book on a shelf

Problem: Find the energy possessed by an object of mass 10 kg when it is at a height of 6 m above the ground. Given g = 9.8 m/s².

Given: m = 10 kg, h = 6 m, g = 9.8 m/s²

Solution: PE = mgh = 10 × 9.8 × 6 = 588 J

NCERT Example 11.6 — Conservation of energy: stone thrown upward

Problem: A stone of mass 0.1 kg is thrown vertically upward with a speed of 10 m/s. Find its (a) initial KE, (b) PE at maximum height, (c) maximum height. Take g = 10 m/s².

Solution:

(a) Initial KE = ½mv² = ½ × 0.1 × 100 = 5 J

(b) At maximum height, v = 0, so all KE converts to PE. PE at max height = 5 J

(c) PE = mgh ⇒ h = PE/(mg) = 5/(0.1 × 10) = 5 m

NCERT Example 11.7 — Power of a person climbing stairs

Problem: A person of mass 50 kg climbs up a staircase of 45 steps in 9 seconds. Each step is 15 cm high. Calculate the power. Take g = 10 m/s².

Given: m = 50 kg, n = 45, h per step = 0.15 m, t = 9 s, g = 10 m/s²

Solution:

Total height H = 45 × 0.15 = 6.75 m

Work done W = mgh = 50 × 10 × 6.75 = 3375 J

Power P = W/t = 3375/9 = 375 W

12. Important solved problems and activities

Activity — Negative work by friction

A block slides on a rough floor. The friction force acts backward (opposite to displacement). Here θ = 180°, so W_friction = Fs cos 180° = −Fs. Friction always does negative work on the sliding object and converts kinetic energy to heat.

Problem — Work done against gravity lifting a suitcase

Problem: Calculate the work required to lift a suitcase of mass 15 kg to a height of 2 m. g = 10 m/s².

Solution: W = mgh = 15 × 10 × 2 = 300 J

Conservation problem — ball falling from 20 m

Problem: A ball of mass 1 kg is dropped from a height of 20 m. Find its KE and PE at heights 20 m, 10 m and 0 m. (g = 10 m/s²)

Total E = mgh = 1 × 10 × 20 = 200 J (constant throughout)

Height PE = mgh KE KE + PE
20 m200 J0 J200 J
10 m100 J100 J200 J
0 m (just before impact)0 J200 J200 J
Power and commercial unit problem

Problem: A household has four 60 W bulbs, one 1000 W heater, and one 100 W fan, all used for 5 hours per day. Find (a) total power, (b) energy consumed per day in kWh, (c) monthly bill at Rs. 5 per unit.

Solution:

(a) Total power = (4 × 60) + 1000 + 100 = 240 + 1000 + 100 = 1340 W

(b) Energy per day = 1340 W × 5 h = 6700 Wh = 6.7 kWh

(c) Monthly energy = 6.7 × 30 = 201 kWh; Bill = 201 × 5 = Rs. 1005

13. Common mistakes and exam tips

  • Work is not always done when force is applied — if there is no displacement, or force is perpendicular to displacement, W = 0.
  • Negative work does not mean no work — friction, air resistance, and brakes do negative work (they remove energy from the system).
  • KE is never negative (v² is always positive); PE can be negative if the reference level is above the object.
  • When using conservation of energy, identify the reference level for PE carefully (usually ground = 0).
  • Power is not energy — a high-power device uses more energy per second, but if used briefly, it may use less total energy than a low-power device used for a long time.
  • In kWh calculations, convert Watts to kilowatts first (divide by 1000) before multiplying by hours.
  • 1 kWh = 3.6 × 10&sup6; J — memorise this conversion exactly; it appears as a 1-mark fill-in frequently.
  • The derivation of KE = ½mv² is a board favourite for 3-mark questions — write all four steps clearly.
Practice MCQs
1. Work done is zero when the angle between force and displacement is:
  1. 45°
  2. 90°
  3. 180°
Answer: (C) 90° — W = Fs cos 90° = 0. Force perpendicular to displacement does no work (e.g., gravity on a horizontally moving object).
2. The kinetic energy of an object of mass m moving with velocity v is:
  1. mv
  2. mv²
  3. ½mv²
  4. 2mv²
Answer: (C) ½mv² — standard NCERT formula derived from work-energy theorem.
3. If the speed of an object is doubled, its kinetic energy becomes:
  1. doubled
  2. halved
  3. quadrupled
  4. unchanged
Answer: (C) quadrupled — KE is proportional to v², so doubling v increases KE by 2² = 4 times.
4. The commercial unit of electrical energy is:
  1. Watt
  2. Joule
  3. kilowatt-hour
  4. kilowatt
Answer: (C) kilowatt-hour (kWh) — also called 1 “unit” of electricity on household bills.
5. 1 kWh equals:
  1. 3.6 × 10³ J
  2. 3.6 × 10&sup4; J
  3. 3.6 × 10&sup5; J
  4. 3.6 × 10&sup6; J
Answer: (D) 3.6 × 10&sup6; J — 1 kWh = 1000 W × 3600 s = 3,600,000 J.
6. A ball of mass 0.5 kg is thrown vertically upward. When it reaches maximum height, its kinetic energy is:
  1. maximum
  2. equal to its initial KE
  3. zero
  4. equal to its weight
Answer: (C) zero — at maximum height velocity = 0, so KE = ½mv² = 0. All energy is stored as PE.
7. Work done by friction on a sliding block is:
  1. positive
  2. zero
  3. negative
  4. equal to the applied force
Answer: (C) negative — friction acts opposite to displacement (θ = 180°), so W = Fs cos 180° = −Fs.
8. A 40 W electric fan is used for 5 hours per day. Energy consumed in 30 days (in kWh) is:
  1. 6 kWh
  2. 0.6 kWh
  3. 60 kWh
  4. 600 kWh
Answer: (A) 6 kWh — E = 40 W × 5 h × 30 days = 6000 Wh = 6 kWh.
9. The gravitational potential energy of an object of mass m at height h is:
  1. mh/g
  2. mg/h
  3. mgh
  4. ½mgh²
Answer: (C) mgh — PE = mgh, with g = 9.8 m/s² and h measured from the chosen reference level.
10. A machine does 1000 J of work in 20 seconds. Its power is:
  1. 20000 W
  2. 200 W
  3. 50 W
  4. 0.02 W
Answer: (C) 50 W — P = W/t = 1000/20 = 50 W.
11. Which of the following is NOT a unit of energy?
  1. Joule
  2. kWh
  3. Watt
  4. erg
Answer: (C) Watt — Watt is the unit of power (energy per unit time), not energy. Joule, kWh, and erg are all units of energy.
12. The work–energy theorem states that the net work done on an object equals:
  1. its kinetic energy
  2. the change in its kinetic energy
  3. the change in its potential energy
  4. its total mechanical energy
Answer: (B) the change in its kinetic energy — W_net = ΔKE = KE_f − KE_i.
Previous-year questions (PYQs)
PYQ 1. Derive the expression for kinetic energy of an object of mass m moving with velocity v. (CBSE Board, 3 marks)
Answer: Start with object at rest (u = 0), mass m, constant force F, displacement s, reaching velocity v. By Newton’s second law F = ma. Work done: W = Fs = mas. From v² = 0 + 2as, we get as = v²/2. Substituting: W = m(v²/2) = ½mv². This work equals the kinetic energy stored: KE = ½mv².
PYQ 2. State the law of conservation of energy. Show that the total mechanical energy of a freely falling body is conserved. (CBSE Board, 5 marks)
Answer: Law: Energy can neither be created nor destroyed; it can only be converted from one form to another. Total energy of an isolated system remains constant. Proof for free fall: Let mass m be dropped from height H. At height H: KE = 0, PE = mgH, Total E = mgH. At height h (less than H): velocity v where v² = 2g(H−h). KE = ½mv² = mg(H−h), PE = mgh, Total E = mg(H−h) + mgh = mgH. At ground (h = 0): KE = mgH, PE = 0, Total E = mgH. Total mechanical energy = mgH at every point — hence conserved.
PYQ 3. A student weighing 50 kg climbs 10 steps of a staircase, each step being 20 cm high, in 20 s. Calculate the work done and the power developed. (g = 10 m/s²) (CBSE Board, 3 marks)
Answer: Total height H = 10 × 0.20 = 2 m. Work W = mgh = 50 × 10 × 2 = 1000 J. Power P = W/t = 1000/20 = 50 W.
PYQ 4. An electric bulb of 60 W is used for 6 hours per day. Calculate the units of energy consumed in one day. What would it cost at Rs. 3 per unit? (CBSE Board, 3 marks)
Answer: Energy = 60 W × 6 h = 360 Wh = 0.36 kWh = 0.36 units. Cost = 0.36 × 3 = Rs. 1.08.
PYQ 5. Define 1 Watt. A device has a power rating of 500 W. How much energy does it transfer in 30 minutes? Express in both Joules and kWh. (CBSE Board, 3 marks)
Answer: 1 Watt is the power of a device that does 1 Joule of work per second (1 W = 1 J/s). Time = 30 min = 1800 s. Energy in Joules = P × t = 500 × 1800 = 9 × 10&sup5; J. In kWh: E = 500 W × 0.5 h = 250 Wh = 0.25 kWh.
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