The Work-energy Theorem for a Variable Force

From Constant Force to Variable Force

The work-energy theorem you already know — $W = \Delta K$ — was proved for a constant force. But forces in nature are rarely constant. A spring's pull gets stronger as you stretch it; the gravitational force on a satellite weakens with distance. The theorem must hold for these cases too, and proving it requires calculus.

The key shift is this: when force varies with position, you cannot simply write $W = F s$. Instead, you must add up the work done over tiny displacements where the force is effectively constant. That sum becomes an integral.

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Proving the Work-Energy Theorem for a Variable Force

We start from Newton's second law: $F = m a = m \frac{dv}{dt}$. But we want to integrate over position, not time. Use the chain rule to rewrite acceleration:

So $F = m v \frac{dv}{dx}$. Now the work integral becomes:

The limits change because when $x = x_i$, $v = v_i$, and when $x = x_f$, $v = v_f$. The integral is elementary:

That is $W = K_f - K_i = \Delta K$. The theorem holds for a variable force exactly as it does for a constant one.

›Proof

Full step-by-step derivation

1. Newton's second law: $F = m a = m \frac{dv}{dt}$.

2. Chain rule: $\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.

3. Hence $F = m v \frac{dv}{dx}$.

4. Work: $W = \int_{x_i}^{x_f} F \, dx = \int_{x_i}^{x_f} m v \frac{dv}{dx} \, dx$.

5. Cancel $dx$: $W = \int_{v_i}^{v_f} m v \, dv$.

6. Integrate: $W = m \left[ \frac{v^2}{2} \right]_{v_i}^{v_f} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 = K_f - K_i$.

The theorem is proved.

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Standard Integrals for Work Calculations

The textbook lists these basic integrals that appear frequently in work-energy problems. You should know them by heart.