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Kinetic Energy and the Work-Energy Theorem

Discover why the energy of motion goes as speed squared, and meet the theorem that turns net work directly into a change in speed.

Energy of motion

A moving object carries energy simply because it is moving — that is its kinetic energy. It depends on the object's mass and, crucially, on the square of its speed.

KE = \tfrac{1}{2}mv^2

Kinetic energy: half the mass times the speed squared.

Deriving the work-energy theorem

Where does \tfrac{1}{2}mv^2 come from? Start from Newton's second law. Suppose a constant net force F acts along the motion over a distance d, giving acceleration a = F/m. The net work is F d = m a d.

W_\text{net} = F d = m a d

Net work written using Newton's second law.

Now borrow a kinematics result from the motion guides: for constant acceleration, v_f^2 = v_i^2 + 2ad. Solve it for ad and substitute.

v_f^2 = v_i^2 + 2 a d \;\Rightarrow\; a d = \tfrac{1}{2}\left(v_f^2 - v_i^2\right)

The constant-acceleration relation rearranged for ad.

W_\text{net} = \tfrac{1}{2}m v_f^2 - \tfrac{1}{2}m v_i^2 = \Delta KE

The work-energy theorem: net work equals the change in kinetic energy.

This is the work-energy theorem. Although we derived it for a constant force, integrating F\,dx shows it holds for any force, constant or varying, straight-line or curved. It is one of the most useful single sentences in mechanics: net work in, change in kinetic energy out.

Reading it as energy bars

As positive net work is done, the kinetic-energy bar grows; negative net work shrinks it. The bar height is literally the running total of work delivered.

The sign of the net work is now easy to interpret. Positive net work → the object speeds up. Negative net work → it slows down. Zero net work → its speed is unchanged, even if it is turning along a curve (as in uniform circular motion, where the net force is perpendicular to the velocity).

Worked example: braking distance

A 1200 kg car travels at 20 m/s when the driver brakes hard. The road provides a constant 6000 N of friction. How far does the car travel before stopping? Strategy: the friction force is the only horizontal force, so its work is the net work; set it equal to the change in kinetic energy (from \tfrac{1}{2}mv_i^2 down to zero).

W_\text{net} = -f\,d = 0 - \tfrac{1}{2}m v_i^2 \;\Rightarrow\; d = \frac{m v_i^2}{2 f}

Set the friction work equal to the loss of kinetic energy and solve for the distance.

d = \frac{(1200)(20)^2}{2(6000)} = 40\ \text{m}

The car needs 40 m to stop — no need to find the acceleration or the time.

Why this is powerful

Notice what the theorem let us skip: acceleration, time, and the detailed shape of the motion. We connected two speeds using only the net work. In the next guide we go one step further — for special 'conservative' forces we can store work as potential energy and never compute work at all.