PhysicsISC Class 11 · Work, Energy and Power

Work-Energy Theorem: Variable Force, Springs, Friction and Power

Updated 17 October 2026 · Fact-checked

The work-energy theorem states that the net work done on a body equals the change in its kinetic energy: W_net = ½mv² − ½mu². Work by a variable force is the integral of F dx (or the area under the F-x graph). In the absence of non-conservative forces, K + U stays constant.

Work by a changing force is the shaded area under the F-x graph, and the net work equals the change in kinetic energy.

Illustration drawn for QuestionBuddy.

Understand Work-Energy Theorem and Variable Force

Work. For a constant force, W = F·s = Fs cosθ. It is zero for a force perpendicular to the displacement (centripetal force, normal reaction on a horizontal motion) and negative when the force opposes the displacement (friction). For a variable force along a line, W = ∫F dx, the area under the force-displacement graph.

Work-energy theorem. The net work of all forces, conservative or not, equals the change in kinetic energy. Use it when the path or the time is not asked, because it avoids the equations of motion.

Potential energy and conservation. Gravitational potential energy is mgh near the Earth's surface. The elastic energy of a spring is ½kx² (compressed or stretched by x), and the work done in changing x from x₁ to x₂ is ½k(x₂² − x₁²). If only conservative forces act, K + U is constant. If friction acts, the loss in mechanical energy equals the work done against friction, and total energy (including heat) is conserved.

Power. P = W/t on average and P = F·v instantaneously. For a vehicle at constant speed on a level road the traction force equals the resistive force, so P = F_resistive × v. Efficiency = (useful output)/(input) × 100%. Units: watt; 1 kW h = 3.6 × 10⁶ J.

Method. Draw the situation, list the forces doing work, find each work (sign included) and equate the sum to ΔK, or use energy conservation with the friction loss added.

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

Work: $W = F scos θ ∣ W = \int F d x$; The integral is the area under the F-x graph.
Work-energy theorem: $W_{n} e t = ½ m v^{2} - ½ m u^{2}$; Net work of all forces.
Spring energy: $U = ½ k x^{2} ∣ W = ½ k (x_{2}^{2} - x_{1}^{2})$; k = force constant.
Power: $P = W / t ∣ P = F v$; Average and instantaneous.

Work-Energy Theorem and Variable Force 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.

Water pump

Suppose a pump lifts 600 kg of water every minute from a well to a tank 20 m higher (g = 10 m s⁻²) and delivers it with negligible speed.

Find the power needed if the pump is 100% efficient, and the power of the motor if its efficiency is 80%.

Try it first, then show the solutionSolution

Work per minute = mgh = 600 × 10 × 20 = 1.2 × 10⁵ J.
Useful power = 1.2 × 10⁵/60 = 2000 W.
Motor power = 2000/0.80 = 2500 W.

Answer: 2000 W useful; 2500 W input.

Memory hook: Input power = output power divided by efficiency.

Tricks to remember Work-Energy Theorem and Variable Force

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

Area gives work
In any F-x graph, area above the axis is positive work and below is negative.
Use it when: Graph-based work questions.
Sign check on friction
Kinetic friction opposing the motion does negative work on a sliding body, so the work is usually negative for a sliding body. Static friction can do positive work (e.g. a box on an accelerating truck, or a person walking).
Use it when: Energy-balance questions with rough surfaces.

Work-Energy Theorem and Variable Force: solved examples

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

Example 1: Change in kinetic energy · 1 mark · MCQBoard level

The speed of a 2 kg body changes from 3 m s⁻¹ to 5 m s⁻¹. The net work done on it is:
(a) 4 J
(b) 8 J
(c) 16 J
(d) 34 J

Try it first, then show the solutionSolution

W = ½m(v² − u²) = ½ × 2 × (25 − 9) = 16 J.

Answer: (c) 16 J

Example 2: Work at an angle · 3 marks · Short answerBoard level

A force of 10 N acting at 60° to the horizontal moves a 2 kg block 5 m along a smooth floor from rest. Find the work done and the final speed.

Try it first, then show the solutionSolution

W = Fs cos60° = 10 × 5 × 0.5 = 25 J. (1 mark)
By the theorem, ½mv² = 25 → v² = 25. (1 mark)
v = 5 m s⁻¹. (1 mark)

Answer: W = 25 J; v = 5 m s⁻¹.

Example 3: Variable force · 4 marks · Short answerHarder board level

A force F = (3x² + 2x) N acts on a 4 kg body along the x-axis, moving it from x = 1 m to x = 3 m. Find the work done and, if the body starts at 2 m s⁻¹, its final speed.

Try it first, then show the solutionSolution

W = ∫₁³ (3x² + 2x) dx = [x³ + x²]₁³. (1 mark)
= (27 + 9) − (1 + 1) = 34 J. (1 mark)
Initial K = ½ × 4 × 4 = 8 J; final K = 8 + 34 = 42 J. (1 mark)
v = √(2 × 42/4) = √21 = 4.58 m s⁻¹. (1 mark)

Answer: W = 34 J; v = 4.58 m s⁻¹.

Example 4: Spring gun · 4 marks · Short answerHarder board level

A spring of force constant 200 N m⁻¹ is compressed by 0.1 m and releases a 50 g ball on a smooth horizontal surface. Find the speed of the ball, and the extra work needed to compress the spring from 0.1 m to 0.2 m.

Try it first, then show the solutionSolution

Energy stored = ½kx² = ½ × 200 × 0.01 = 1 J. (1 mark)
½mv² = 1 → v = √(2 × 1/0.05) = √40 = 6.32 m s⁻¹. (1 mark)
Extra work = ½k(0.2² − 0.1²) = 100 × 0.03 = 3 J. (2 marks)

Answer: v = 6.32 m s⁻¹; extra work 3 J.

Example 5: Incline then rough floor · 5 marks · Long answerHOTS / competency

A 2 kg block is released from rest at the top of a 5 m long incline at 30° to the horizontal, with μ = 0.2 on the incline (g = 10 m s⁻²). It then slides on a rough horizontal floor with the same μ. Find its speed at the foot of the incline and how far it travels on the floor before stopping.

Try it first, then show the solutionSolution

Work by gravity = mg sin30° × 5 = 2 × 10 × 0.5 × 5 = 50 J. (1 mark)
Work against friction = μ mg cos30° × 5 = 0.2 × 20 × 0.866 × 5 = 17.3 J. (1 mark)
K at the foot = 50 − 17.3 = 32.7 J; v = √(2 × 32.7/2) = 5.72 m s⁻¹. (1 mark)
On the floor: friction μmg = 0.2 × 20 = 4 N. (1 mark)
Distance = 32.7/4 = 8.17 m. (1 mark)

Answer: v = 5.72 m s⁻¹ at the foot; stops after 8.2 m on the floor.

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

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

Mistake: Forgetting the sign of friction work
Adding the friction work instead of subtracting.
Do this: Write W_friction as negative and show it in the energy equation.
Usually costs: 1 to 2 marks.
Mistake: Using a constant-force formula for a variable force
Taking F × s with F at one point.
Do this: Integrate F dx between the given limits.
Usually costs: 2 marks.
Mistake: Not converting units
Using grams or kilowatts directly.
Do this: Convert to kg, m, s, J, W before substituting.
Usually costs: 1 mark.

Exam tips

Name the system and list every force that does work, with its sign.
Use the work-energy theorem when time and path are not required.
Show the integral or the area calculation explicitly for variable forces.
Give units: J for work and W for power.

Practice questions on Work-Energy Theorem and Variable Force

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 · MCQBoard level
A block slides a distance s along a horizontal table. The work done on it by the normal reaction N is:
(a) Ns
(b) −Ns
(c) zero
(d) mgs
Hint
N is perpendicular to the displacement.
Answer
(c) zero
Q2 · 2 marks · Short answerBoard level
A crane lifts a 500 kg load through 12 m in 20 s at steady speed (g = 10 m s⁻²). Find the power it delivers to the load.
Hint
P = mgh/t.
Answer
P = 500 × 10 × 12/20 = 3000 W.
Q3 · 3 marks · Short answerHarder board level
A 0.2 kg ball dropped from 20 m rebounds to 12 m (g = 10 m s⁻²). Find the loss of mechanical energy and the percentage lost.
Hint
Compare mgh before and after.
Answer
Loss = 0.2 × 10 × (20 − 12) = 16 J. Initial energy = 40 J, so the loss is 40%.
Q4 · 3 marks · Short answerHarder board level
A force rises uniformly from 0 to 20 N over the first 4 m of displacement and then stays at 20 N for the next 6 m. Find the total work done.
Hint
Area under the F-x graph.
Answer
W = ½ × 4 × 20 + 6 × 20 = 40 + 120 = 160 J.
Q5 · 4 marks · Short answerHarder board level
A 1000 kg car has an engine of 40 kW and a top speed of 40 m s⁻¹ on a level road. Find the resistive force, and the top speed up an incline of slope 1 in 20 assuming the same resistive force (g = 10 m s⁻²).
Hint
At top speed, traction equals the total opposing force.
Answer
F_res = P/v = 40000/40 = 1000 N. On the slope the opposing force = 1000 + 1000 × 10/20 = 1500 N, so v = 40000/1500 = 26.7 m s⁻¹.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a frictionless roller coaster car of mass 300 kg is released from rest at a height of 25 m and goes round a vertical loop of radius 8 m (g = 10 m s⁻²).
(i) Find its speed at a height of 5 m. (1 mark)
(ii) Find its speed at the bottom. (1 mark)
(iii) Find the minimum speed needed at the top of the loop. (1 mark)
(iv) Find its speed at the top of the loop (height 16 m) and say whether it stays on the track. (1 mark)
Hint
Use mgh lost = ½mv² gained; the top of the loop needs v² ≥ gr.
Answer
(i) v = √(2 × 10 × 20) = 20 m s⁻¹. (ii) √(2 × 10 × 25) = 22.4 m s⁻¹. (iii) √(gr) = √80 = 8.94 m s⁻¹. (iv) v² = 2 × 10 × (25 − 16) = 180, v = 13.4 m s⁻¹ > 8.94 m s⁻¹, so it stays on the track.

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.State the work-energy theorem.Show answer
Net work done on a body equals the change in its kinetic energy.
2.Energy stored in a spring?Show answer
½kx².
3.Instantaneous power?Show answer
P = F·v.
4.How is 1 kW h related to joules?Show answer
1 kW h = 3.6 × 10⁶ J.

Work-Energy Theorem and Variable Force: frequently asked questions

Does the work-energy theorem work when friction acts?

Yes. The net work includes the negative work of friction. The theorem holds for all forces; only the conservation of mechanical energy needs conservative forces.

Can work be negative?

Yes. When the force has a component opposite to the displacement (for example friction, or gravity on a rising body), the work is negative.

Is energy lost when friction acts?

Mechanical energy is converted to heat and sound; the total energy is conserved.