Work, Energy & Power

Work by a Variable Force

$W = \int_{x_i}^{x_f} F(x)\,dx$

For a constant force: $W = Fd\cos\theta$.

For a spring: $W = \int_0^x (-kx')\,dx' = -\frac{1}{2}kx^2$.

Work-Energy Theorem

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

Conservative Forces and Potential Energy

A force is conservative if:

• Work depends only on endpoints, not path.
• Work around any closed loop is zero.

$F(x) = -\frac{dU}{dx}$

Examples:

• Gravity near surface: $U = mgy$ → $F = -mg$ (downward).
• Spring: $U = \frac{1}{2}kx^2$ → $F = -kx$.
• Universal gravity: $U = -\frac{GMm}{r}$ → $F = -\frac{GMm}{r^2}$ (attractive).

Conservation of Mechanical Energy

With only conservative forces: $E = KE + U = \text{constant}$

With non-conservative forces: $\Delta E = W_{\text{nc}}$

Power

$P = \frac{dW}{dt} = \vec{F} \cdot \vec{v}$

Problem: A force $F(x) = 5x^2$ N acts on a $3\ \text{kg}$ object from $x = 0$ to $x = 2\ \text{m}$ (starting from rest). Find the final speed.

Solution:

$W = \int_0^2 5x^2\,dx = \frac{5x^3}{3}\Big|_0^2 = \frac{5(8)}{3} = \frac{40}{3}\ \text{J}$

By the work-energy theorem: $\frac{40}{3} = \frac{1}{2}(3)v^2$ $v = \sqrt{\frac{80}{9}} \approx 2.98\ \text{m/s}$

• Work by a variable force: $W = \int F\,dx$.
• $F = -dU/dx$ connects force and potential energy.
• Conservation of energy: $KE + U = \text{const}$ when only conservative forces act.
• Power: $P = dW/dt = \vec{F} \cdot \vec{v}$.