Kinetic Energy and the Work-Energy Theorem

A moving body has the capacity to do work — push it into another body and it can dent, lift, or break things. We call this stored capacity its kinetic energy.

Concept

The kinetic energy of a particle of mass $m$ moving with speed $v$ is

$K = \tfrac{1}{2}mv^2$

It is a scalar and is always non-negative. SI unit: joule.

In terms of momentum $p = mv$,

$K = \frac{p^2}{2m}$

Work-Energy Theorem (WKT): The net work done by all forces on a particle equals the change in its kinetic energy.

$W_{\text{net}} = \Delta K = K_f - K_i = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$

This theorem holds in any inertial frame, for constant or variable forces, conservative or not.

Derivation

Start with Newton's second law in 1D for a particle:

$F = m\frac{dv}{dt} = m\frac{dv}{dx}\cdot\frac{dx}{dt} = mv\frac{dv}{dx}$

Multiplying by $dx$ and integrating from initial position $x_i$ to final $x_f$:

$\int_{x_i}^{x_f} F\,dx = \int_{v_i}^{v_f} mv\,dv$

The left side is $W_{\text{net}}$ (work done by net force). The right side evaluates to

$\tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 = \Delta K$

Hence

$W_{\text{net}} = \Delta K$

In 3D, the same argument applies using $\vec{F}\cdot\vec{v} = m\,\vec{v}\cdot d\vec{v}/dt = d(\tfrac{1}{2}mv^2)/dt$.

Worked Example

A 2 kg block, initially at rest, is pushed along a horizontal floor by a constant horizontal force of 10 N over 5 m. Kinetic friction is 4 N. Find (a) the final speed using WKT, (b) verify using kinematics.

Solution:

(a) Net force: $F_{\text{net}} = 10 - 4 = 6 \, \text{N}$.

Work by net force: $W_{\text{net}} = 6 \times 5 = 30 \, \text{J}$.

By WKT: $\Delta K = W_{\text{net}} = 30 \, \text{J}$. Starting from rest:

$\tfrac{1}{2}(2)v_f^2 = 30 \Rightarrow v_f^2 = 30 \Rightarrow v_f = \sqrt{30} \approx 5.48 \, \text{m/s}$

(b) Acceleration: $a = 6/2 = 3 \, \text{m/s}^2$. $v_f^2 = 0 + 2(3)(5) = 30$. Same answer.

Common Confusions

• WKT uses net work, not just one force's work.
• KE depends on frame. A book at rest in a moving train has KE in the ground frame.
• KE is always non-negative, even if velocity has a "negative direction" — because $v^2$ is positive.
• Sign of $\Delta K$. Negative net work decreases KE (e.g., a braking car).

Key Takeaways

• $K = \tfrac{1}{2}mv^2 = p^2/(2m)$, always non-negative.
• Work-energy theorem: $W_{\text{net}} = \Delta K$.
• Holds for all force types — constant, variable, conservative, non-conservative.
• Useful shortcut when forces are complicated but the start and end speeds are wanted.
• KE depends on the frame of reference.