Work-Energy Theorem

The theorem

The net work done on a body by all forces equals the change in its kinetic energy:

$W_{net} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

This is one of the most powerful results in mechanics. It directly connects force (through work) to motion (through kinetic energy), without requiring us to know the details of how the velocity changed at every instant.

Derivation

For a body of mass $m$ moving in one dimension under net force $F_{net}$:

By Newton's Second Law: $F_{net} = ma$

Using the kinematic identity $a = v\frac{dv}{dx}$:

$F_{net} = mv\frac{dv}{dx}$

Work done by net force over displacement from $x_1$ to $x_2$:

$W_{net} = \int_{x_1}^{x_2} F_{net} \, dx = \int_{x_1}^{x_2} mv\frac{dv}{dx} \, dx = \int_{v_i}^{v_f} mv \, dv$

$W_{net} = m \cdot \frac{v^2}{2}\Big|_{v_i}^{v_f} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

$\boxed{W_{net} = \Delta KE}$

Why this is powerful

The work-energy theorem bypasses the need to solve $F = ma$ as a differential equation. Instead of tracking velocity at every instant, you just need:

• The net work done (which may be found from force-displacement graphs or direct calculation)
• The initial and final speeds

Example: A 2 kg block starts at rest and is pushed by a net force along a 5 m surface. If the net work done is 40 J, find the final speed.

$40 = \frac{1}{2}(2)v_f^2 - 0 \implies v_f^2 = 40 \implies v_f = \sqrt{40} \approx 6.3 \text{ m/s}$

No need to find acceleration or use equations of motion.

Net work — all forces included

$W_{net}$ includes work done by every force: applied force, friction, gravity, normal force, spring, etc.

• Forces perpendicular to motion (normal force, centripetal force) do zero work
• Friction does negative work
• Gravity does $mgh$ (positive downward, negative upward)

Special cases

Body in equilibrium ($v_f = v_i$): $W_{net} = 0$ — net work is zero when speed does not change.

Body decelerating to rest ($v_f = 0$): $W_{net} = -\frac{1}{2}mv_i^2$ — negative net work removes kinetic energy.

Free fall from rest through height $h$: $W_{gravity} = mgh = \frac{1}{2}mv_f^2$, giving $v_f = \sqrt{2gh}$ — consistent with kinematics.

Relation to Newton's Second Law

The work-energy theorem is not independent of Newton's Second Law — it is derived from it. It is a useful restatement that is particularly convenient for problems involving curved paths or variable forces, where direct application of $F = ma$ would be complicated.