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Work Energy Theorem Formula

Work, energy and collision formulasClass 11AP Physics 1Updated 1 Nov 2026

Wnet=12mv2−12mu2W_{net} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2
SI unit
J (joule)
Dimensions
[M¹ L² T⁻²]
where
  • W_net net work done by all forces J
  • m mass kg
  • v final speed m/s
  • u initial speed (vᵢ or v₀ in US books) m/s

The work energy theorem formula is Wnet=12mv2−12mu2W_{net} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2: the total work done on a body by all the forces acting on it equals the change in its kinetic energy. Positive net work speeds a body up, negative net work slows it down, and zero net work leaves its speed unchanged.

It is the shortcut that lets you skip acceleration and time. If you know the forces and the distance, you get the speed straight away, even when the path curves. Here u is the starting speed and v the final one, as in Indian textbooks; US books often use vᵢ and v_f. The theorem is one of the main results in the work and energy chapter.

Quick facts

  • Formula

    Wnet=ΔKEW_{net} = \Delta KE

  • Expanded

    Wnet=12mv2−12mu2W_{net} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

  • SI unit

    joule (J)

  • Dimensional formula

    [M¹ L² T⁻²]

  • Net work

    Sum of the work by every force, including friction and gravity

  • Valid for

    Constant or varying forces, straight or curved paths

  • Courses

    Class 11, AP Physics 1 and C, JEE, NEET

Reading the work energy theorem formula

Parts of the theorem
TermMeaningUnit
WnetW_{net}work done by the resultant force, or the sum of the works of all forcesJ
12mu2\frac{1}{2}mu^2kinetic energy the body has before any workJ
12mv2\frac{1}{2}mv^2kinetic energy once the work is doneJ

The word net is the key. A crate pulled at steady speed has positive work from the rope and negative work from friction, and the two add to zero, so its kinetic energy does not change. Work done by a single force is not enough unless it is the only force doing work.

Proof for a constant force

  1. A net force F acts along the motion over a displacement s, so Wnet=FsW_{net} = Fs.
  2. Newton's second law gives F=maF = ma.
  3. The third equation of motion gives v2=u2+2asv^2 = u^2 + 2as, so as=v2−u22as = \frac{v^2 - u^2}{2}.
  4. Then Wnet=mas=12mv2−12mu2W_{net} = mas = \frac{1}{2}mv^2 - \frac{1}{2}mu^2.

For a varying force, write F=mdvdt=mvdvdxF = m\frac{dv}{dt} = mv\frac{dv}{dx} and integrate: ∫F dx=∫uvmv dv=12mv2−12mu2\int F\,dx = \int_u^v mv\,dv = \frac{1}{2}mv^2 - \frac{1}{2}mu^2. This is the version AP Physics C and JEE expect.

Worked examples

Example 1: pushing a toy car

A 2.0 kg toy car rolling at 3.0 m/s is pushed until it reaches 7.0 m/s. How much net work was done on it?

Given: m = 2.0 kg, u = 3.0 m/s, v = 7.0 m/s.

Wnet=12(2.0)(7.02−3.02)=1.0×(49−9)=40 JW_{net} = \frac{1}{2}(2.0)(7.0^2 - 3.0^2) = 1.0 \times (49 - 9) = 40\ \text{J}

Answer: 40 J of net work.

Example 2: braking a car (exam style, km/h)

A 1500 kg car moving at 90 km/h brakes to a stop in 50 m. Find the average braking force.

Convert: 90 km/h ÷ 3.6 = 25 m/s. Final speed v = 0.

Wnet=0−12(1500)(25)2=−468 750 JW_{net} = 0 - \frac{1}{2}(1500)(25)^2 = -468\,750\ \text{J}

The braking force F acts backwards over 50 m, so −F(50)=−468 750-F(50) = -468\,750 and F=9375 N≈9.4 kNF = 9375\ \text{N} \approx 9.4\ \text{kN}.

Answer: about 9.4 kN. At double the speed the car would need four times the distance for the same force.

Write the initial kinetic energy with the right speed. Squaring 90 instead of 25 is the most common slip in this question.

When to use energy instead of forces

Use the theorem when the question gives distances and speeds but not times, or when the path curves, as on a slide or a pendulum. Use Newton's laws when you need the acceleration or the time. In the calculator above, choose which of W, m, v or u is missing. If the net work is negative and larger than the starting kinetic energy, there is no real final speed: the body stops before covering the distance.

When gravity or a spring does the work, you can move that work to the other side as a change in potential energy. That turns the theorem into the conservation of energy formula, which you can use when only conservative forces act.

Frequently asked questions

What is the work energy theorem formula?

W_net = ½mv² − ½mu², or W_net = ΔKE. The net work done on a body equals the change in its kinetic energy.

Does the work energy theorem work for variable forces?

Yes. The proof with integration shows it holds for any force and any path, as long as you use the total work done by all forces.

What does negative net work mean?

The body loses kinetic energy and slows down. Braking and friction are the usual examples.

Is the work energy theorem the same as conservation of energy?

Not exactly. The theorem deals with kinetic energy and the work of all forces. Conservation of mechanical energy follows from it when the work by gravity and springs is written as a change in potential energy and no friction acts.

What is the unit of work and kinetic energy in the theorem?

Both are measured in joules, with dimensions [M¹ L² T⁻²]. That is why the two sides of the equation can be equal.