Work Energy Theorem: Choose the Method, Then Solve the Numerical

The work energy theorem states that the net work done on a body equals the change in its kinetic energy: W_net = ½mv² − ½mu². Here m is the mass, u the starting speed and v the final speed. Net work is the sum of the work done by every force acting on the body. Both quantities are measured in joules (J).

The angle θ between force and displacement sets the sign of the work: at 0° the work is positive, at 90° it is zero, and at 180° it is negative.

Why it is true for a constant force: start from v² = u² + 2as and multiply both sides by m/2. You get mas = ½mv² − ½mu². Since ma is the net force, mas is the net work.

For a force that changes with position, you add up the work over tiny steps. That sum is the integral ∫F dx, which is the area under the force–displacement graph. OpenStax University Physics Volume 1, section 7.3 derives this in full. It also makes the point that matters most in practice: you must include every force that acts on the object, and nothing that does not.

Here is the short version of that derivation for a force that changes with position. Write the force as F = m(dv/dt). Because dv/dt = (dv/dx)(dx/dt) = v(dv/dx), the force is F = mv(dv/dx). Then W = ∫F dx = ∫mv dv, taken from speed u to speed v, which equals ½mv² − ½mu². The result does not depend on how the force varies, which is why the theorem works for springs and graphs.

Use the work energy theorem when the question gives speeds and distances or heights and does not ask for time. Use F = ma when you need acceleration or time. Use conservation of mechanical energy only when gravity and springs are the sole forces doing work. This table is our recommended way to choose.

Follow four steps: mark the start and end, list each force's signed work, add them and equate to the change in kinetic energy, then check sign and units. This is the same habit we teach in How to Solve Physics Numericals, applied to this chapter.

How far does a vehicle slide at 60, 80 and 100 km/h?

A locked-wheel slide takes 20.2 m from 60 km/h, 36.0 m from 80 km/h and 56.2 m from 100 km/h on a level road. This assumes a friction coefficient μ of 0.7, a value we assume for a dry road and not a measured one.

The working goes like this. Gravity and the normal force are perpendicular to the motion and do no work. Friction is f = μmg at 180° to the motion, so its work is −μmg × d. The vehicle ends at rest, so:

−μmg × d = 0 − ½mu², which gives d = u² / (2μg)

The mass cancels, so it does not matter how heavy the vehicle is. We used g = 9.8 m/s² and converted speeds to m/s.

Going from 60 to 80 km/h multiplies the speed by 1.33 and the distance by 1.78, which is 1.33 squared. Going from 60 to 100 km/h multiplies the speed by 1.67 and the distance by 2.78. The distance depends on u², so the speed term is squared, not doubled. This covers only the slide itself, not the distance covered while the driver reacts.

The u² term is not only a classroom idea. According to the Ministry of Road Transport and Highways' Road Accidents in India 2024 report, over-speeding was the traffic violation associated with the largest share of road deaths in 2024: 70.28% of the 1,77,175 people killed in accidents reported by the police departments of States and Union Territories.

How fast does a block reach the bottom of a rough incline?

A 2 kg block starts from rest and slides 5 m down a 30° incline with μ = 0.2. Find its speed at the bottom. We used g = 9.8 m/s² and ignored air resistance.

• The block drops by h = 5 × sin 30° = 2.5 m, so gravity does +mgh = 2 × 9.8 × 2.5 = 49 J.
• The normal force is perpendicular to the motion, so it does zero work.
• Friction is f = μmg cos 30° = 0.2 × 2 × 9.8 × 0.866 = 3.395 N. Over 5 m it does −3.395 × 5 = −16.97 J.
• Net work = 49 − 16.97 = 32.03 J, and the block starts from rest, so 32.03 = ½ × 2 × v².

That gives v² = 32.03 and v = 5.66 m/s. The mass cancels again, so v = √[2gs(sin θ − μ cos θ)] works for any block on this slope.

If your question has friction on an incline or a ladder, our page on friction in engineering mechanics sets up those forces.

The theorem handles a spring or a variable force by replacing "force × distance" with the area under the force–displacement graph. A spring compressed by x stores ½kx², where k is the spring constant. For a force F(x), the work is ∫F dx. Two worked numericals follow, again with our own numbers.

Spring launching a sphere. A 0.1 kg sphere sits on a spring with k = 200 N/m, compressed by 0.05 m. It is released, leaves the spring and rises until it stops. Find the height h above the release point.

• The spring does +½kx² = ½ × 200 × 0.05² = 0.25 J while it unwinds.
• Gravity does −mgh = −0.1 × 9.8 × h over the whole rise.
• The sphere starts and ends at rest, so the net work is zero: 0.25 − 0.98h = 0.

That gives h = 0.255 m, about 0.26 m. We ignored air resistance.

Variable force. A force F = 4x newtons (x in metres) acts along a smooth surface on a 2 kg block starting from rest, from x = 0 to x = 3 m.

• W = ∫ 4x dx from 0 to 3 = 2x² evaluated between 0 and 3 = 18 J.
• Graph check: F rises from 0 to 12 N, so the triangle's area is ½ × 3 × 12 = 18 J.
• 18 = ½ × 2 × v², so v² = 18 and v = 4.24 m/s.

Five mistakes cost the most marks. Each comes from treating a sign or a force carelessly, not from the maths. Run through the list before you hand in the answer.

For the units side of checking an answer, see Units and Dimensions in Class 11 Physics.

When two bodies have the same kinetic energy, the heavier one has more momentum, and the ratio of their momenta is the square root of the ratio of their masses. This follows from writing kinetic energy as E = p²/(2m), so p = √(2mE).

Take masses of 1 g and 4 g with equal kinetic energy. Then p₁ / p₂ = √(1/4) = 1/2, so the momenta are in the ratio 1 : 2. The units of mass cancel in the ratio.

The theorem links to this directly. Two bodies of different mass pushed from rest by the same force over the same distance receive the same net work, so they gain the same kinetic energy. The heavier one ends up slower but carries more momentum. The same force acting for the same time would instead give them equal momentum, and then the lighter one carries more kinetic energy.

If you are sitting MHT-CET, the official State CET Cell syllabus page lists the topics for your paper.

Which work energy numericals should you try yourself?

Try these two before you read the answers. Each one uses the four steps above, and both use our own numbers.

1. A 2 kg block moving at 6 m/s on a rough horizontal floor stops after 4.5 m. Find the friction force and the coefficient of friction μ. Use g = 9.8 m/s².
2. A 0.2 kg ball is dropped from 5 m above soft sand and sinks 0.10 m before it stops. Find the average resistive force from the sand.

Answers: (1) The change in kinetic energy is 0 − ½ × 2 × 6² = −36 J, and friction is the only force doing work, so −f × 4.5 = −36. That gives f = 8 N and μ = 8 / (2 × 9.8) = 0.41. (2) Gravity does +mg × (5 + 0.10) over the whole fall, the sand does −F × 0.10, and the ball starts and ends at rest. So 0.2 × 9.8 × 5.10 − F × 0.10 = 0, which gives F = 99.96 N, about 100 N.

For past-paper practice by chapter, use our MHT-CET PYQ practice library.

We teach numericals from first principles: choose the method, list the forces with their signs, then solve. That is the routine on this page. The Padhna Aasaan (studying made easy) idea is that once the setup is a habit, the chapter stops being a list of formulas to memorise.

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