PhysicsCBSE Class 11 · Work, Energy and Power

Work-Energy Theorem, Variable Force and Power Class 11

Updated 17 October 2026 · Fact-checked

Work W = F·d = Fd cos θ; for a variable force W = ∫F dx, the area under the F-x graph. The work-energy theorem says the net work on a body equals its change in kinetic energy, W = ½mv² − ½mu². Power is the rate of work: P = dW/dt = F·v, and the useful output over the input gives efficiency.

Work is the area under the F-x curve, and the net work equals the change in kinetic energy; power is F times v.

Illustration drawn for QuestionBuddy.

Understand Work-Energy Theorem, Variable Force and Power

Work. For a constant force acting through a displacement $d$ , $W = F \cdot d = F d cos θ$ . Work is a scalar, positive for $θ < 9 0^{\circ}$ , zero at $9 0^{\circ}$ (e.g. centripetal force, normal reaction) and negative for $θ > 9 0^{\circ}$ (friction opposing motion). For a variable force along a line, $W = \int_{x_{1}}^{x_{2}} F d x$ , the area under the F-x graph.

Kinetic energy and the work-energy theorem. $K = \frac{1}{2} m v^{2} = \frac{p ^{2}}{2 m}$ . The work-energy theorem states that the work done by the *net* force equals the change in kinetic energy: $W_{n e t} = K_{f} - K_{i}$ . It holds for any force (conservative or not), so it handles friction and variable forces directly; add the work of every force, including negative work by friction, before applying it.

Incline problems. For a block of mass $m$ moving a distance $L$ down a rough incline of angle $θ$ : $W_{g r a v i t y} = m g L sin θ$ and $W_{f r i c t i o n} = - μ m g L cos θ$ . If the block starts from rest, $\frac{1}{2} m v^{2} = m g L (sin θ - μ cos θ)$ . A body on a rough horizontal surface with kinetic energy $K$ stops after $d = K / (μ m g)$ .

Power and efficiency. Average power $P = W / t$ ; instantaneous power $P = F \cdot v$ . For a machine, efficiency $η = P_{o u t} / P_{in}$ . A pump raising mass $\overset{m}{˙}$ per second through height $h$ and giving it speed $v$ must supply $P = \overset{m}{˙} (g h + \frac{1}{2} v^{2})$ . For a vehicle at constant power, $F = P / v$ and the maximum speed occurs when the driving force equals the resistance. Units: watt (W), kilowatt hour = $3.6 \times 1 0^{6}$ J.

Common traps. A force perpendicular to motion does no work; work by static friction can be non-zero (it can accelerate a body relative to the ground); and a body's kinetic energy is quadrupled when its speed is doubled while its momentum only doubles.

Work-Energy Theorem, Variable Force and Power: formulas and key facts

Work: $W = F d cos θ$; Constant force.
Variable force: $W = \int F d x$; Area under F-x graph.
Kinetic energy: $K = \frac{1}{2} m v^{2} = \frac{p ^{2}}{2 m}$; Scalar.
Work-energy theorem: $W_{n e t} = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$; All forces.
Power: $P = \frac{d W}{d t} = F \cdot v$; Watt.
Efficiency: $η = \frac{P _{o u t}}{P _{in}}$; Less than 1.

How to solve Work-Energy Theorem, Variable Force and Power questions

List the work done by each force with its sign, add them, and set the sum equal to the change in kinetic energy.

1Draw the body and list every force acting while it moves.
2Find the work of each force: Fd cos θ for constant forces, the integral for variable ones.
3Give friction a negative sign and perpendicular forces zero.
4Add them to get W_net and set W_net = ΔK.
5Solve for the unknown speed, distance or force.
6For power, use P = W/t or P = Fv, and apply efficiency at the end.

Work-Energy Theorem, Variable Force and Power in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Stopping a loaded trolley

Suppose a trolley and load of total mass 40 kg moving at 3 m/s on a level floor are brought to rest by friction, μ = 0.15 (g = 10 m/s²).

How far does it slide?

Try it first, then show the solutionSolution

K = ½ × 40 × 9 = 180 J.
Friction force = 0.15 × 400 = 60 N; d = 180/60 = 3 m.

Answer: 3 m.

Memory hook: Stopping work equals the kinetic energy.

An electric hoist

Suppose a hoist lifts a 150 kg load through 12 m in 30 s at steady speed (g = 10 m/s²) and the motor is 75% efficient.

Find the input power drawn.

Try it first, then show the solutionSolution

Work = mgh = 150 × 10 × 12 = 18000 J.
Output power = 18000/30 = 600 W; input = 600/0.75 = 800 W.

Answer: 800 W.

Memory hook: Efficiency divides the output power.

Tricks to remember Work-Energy Theorem, Variable Force and Power

Short memory aids. Say each one aloud once, then use it on a question the same day.

Net work, not each work
The theorem uses the sum of all works, so add friction as negative.
Use it when: Rough-surface problems.
Speed doubles, KE ×4
K ∝ v² and p ∝ v.
Use it when: Ratio questions.
Power in a vehicle
P = Fv, so at maximum speed F = resistance.
Use it when: Engine problems.
Area under F-x
Triangle ½bh, rectangle bh.
Use it when: Graph questions.

Work-Energy Theorem, Variable Force and Power: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Work to change speed · 2 marks · Very short answerBoard level

Find the work required to increase the speed of a 1200 kg car from 10 m/s to 20 m/s.

Try it first, then show the solutionSolution

W = ΔK = ½ × 1200 × (20² − 10²). (1)
= 600 × 300 = 1.8 × 10⁵ J. (1)

Answer: 1.8 × 10⁵ J.

Example 2: Force at an angle with friction · 3 marks · Short answerBoard level

A 30 kg crate is pushed 8 m along a rough level floor by a 100 N force directed 30° below the horizontal. The friction force is 40 N. It starts from rest. Find the work done by the push, the work done by friction, and the final speed.

Try it first, then show the solutionSolution

Work by push = 100 × 8 × cos 30° = 692.8 J. (1)
Work by friction = −40 × 8 = −320 J; net work = 372.8 J. (1)
½ × 30 × v² = 372.8, so v ≈ 5.0 m/s. (1)

Answer: 692.8 J; −320 J; about 5.0 m/s.

Example 3: Variable force · 3 marks · Short answerHarder board level

The force on a 2 kg body at rest is F = 4x + 3x² (N). Find the work done between x = 0 and x = 2 m and the speed at x = 2 m.

Try it first, then show the solutionSolution

W = ∫(4x + 3x²)dx = [2x² + x³] from 0 to 2. (1)
= 8 + 8 = 16 J. (1)
½ × 2 × v² = 16, so v = 4 m/s. (1)

Answer: 16 J; 4 m/s.

Example 4: Incline then level ground · 4 marks · Long answerHarder board level

A 5 kg block starts from rest and slides 6 m down a 37° incline (sin 37° = 0.6, cos 37° = 0.8) with μk = 0.25 (g = 10 m/s²). It then enters a level rough floor with the same μk. Find its speed at the bottom of the incline and how far it travels on the level floor before stopping.

Try it first, then show the solutionSolution

Work by gravity = 5 × 10 × 0.6 × 6 = 180 J. (1)
Work by friction = −0.25 × 50 × 0.8 × 6 = −60 J, so net = 120 J. (1)
½ × 5 × v² = 120 gives v = √48 ≈ 6.9 m/s. (1)
On the floor 120 J = 0.25 × 50 × d, so d = 9.6 m. (1)

Answer: 6.9 m/s; 9.6 m.

Example 5: Pumping water · 4 marks · Long answerHOTS / competency

A pump lifts 1800 kg of water per minute from a well 20 m deep and delivers it at a speed of 10 m/s (g = 10 m/s²). Find the power delivered to the water and the power needed from the motor if the pump is 60% efficient.

Try it first, then show the solutionSolution

Mass flow = 1800/60 = 30 kg/s. (1)
Energy per kg = gh + ½v² = 200 + 50 = 250 J/kg. (1)
Output power = 30 × 250 = 7500 W. (1)
Input power = 7500/0.6 = 12500 W = 12.5 kW. (1)

Answer: 7.5 kW delivered; 12.5 kW input.

Where students lose marks in Work-Energy Theorem, Variable Force and Power

The slips examiners see most often, and what to write instead.

Mistake: Ignoring friction's negative work in the theorem.
Using only the applied force.
Do this: Add the work of all forces with signs.
Usually costs: The answer mark.
Mistake: Using Fd instead of Fd cos θ.
Forgetting the angle.
Do this: Resolve along the displacement.
Usually costs: 1 mark.
Mistake: Treating power as energy.
Mixing watts and joules.
Do this: P = W/t; E = Pt.
Usually costs: Unit mark.
Mistake: Adding efficiency incorrectly (multiplying instead of dividing for input).
Mixing output and input.
Do this: Input = output ÷ efficiency.
Usually costs: The answer mark.

Exam tips

Write "W_net = ΔK" first and list the works below it.
Do not compute the acceleration if the work-energy theorem gives the speed directly.
State the sign of each work term.
For variable forces draw the F-x graph and use areas where possible.
Convert units (km/h, kWh) before substituting.

Practice questions on Work-Energy Theorem, Variable Force and Power

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQHarder board level
The kinetic energy of a body increases by 300%. Its momentum increases by
(a) 100% (b) 200% (c) 300% (d) 400%
Hint
p ∝ √K.
Answer
(a) 100%: K becomes 4 times, so p becomes 2 times.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The work done by static friction on a body can be non-zero.
Reason (R): Static friction can act on a body that is moving relative to the ground, such as a box carried on an accelerating truck.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Look at the displacement relative to the ground.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
A motor lifts a 50 kg load at a steady 0.4 m/s. Find the power it supplies to the load (g = 10 m/s²).
Hint
P = Fv.
Answer
P = 500 × 0.4 = 200 W.
Q4 · 3 marks · Short answerHarder board level
A 0.2 kg mass is dropped from 0.5 m above a vertical spring of constant 100 N/m. Find the maximum compression of the spring (g = 10 m/s²).
Hint
mg(h + x) = ½kx².
Answer
2(0.5 + x) = 50x² gives 50x² − 2x − 1 = 0, so x = (2 + √204)/100 ≈ 0.163 m.
Q5 · 4 marks · Long answerHOTS / competency
A 1000 kg car has an engine of constant power 40 kW and the resistive force is constant at 1000 N. Find (a) its maximum speed, and its acceleration at (b) 10 m/s and (c) 20 m/s. Why does the acceleration fall with speed?
Hint
Driving force F = P/v.
Answer
(a) vmax = P/f = 40000/1000 = 40 m/s. (b) F = 4000 N, a = (4000 − 1000)/1000 = 3 m/s². (c) F = 2000 N, a = 1 m/s². At constant power the driving force P/v falls as v rises, so the net force falls.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Work-energy theorem?Show answer
W_net = ΔK.
2.Work by a centripetal force?Show answer
Zero.
3.Power in terms of force and velocity?Show answer
P = F·v.
4.1 kWh in joules?Show answer
3.6 × 10⁶ J.
5.Work from an F-x graph?Show answer
Area under the curve.
6.KE in terms of momentum?Show answer
p²/2m.

Work-Energy Theorem, Variable Force and Power: frequently asked questions

Does the theorem apply to non-conservative forces?

Yes. It applies to the net work of every force, conservative or not.

Can work be negative?

Yes, when the force opposes the displacement, as friction does.

Why does a car have a maximum speed at constant power?

Because the driving force P/v falls with speed until it equals the resistance.

How is the kilowatt hour a unit of energy?

1 kW for 1 hour is 1000 W × 3600 s = 3.6 × 10⁶ J.

Is kinetic energy ever negative?

No, since it depends on v².