Constant force:
$$W = \mathbf{F}\cdot\mathbf{s} = Fs\cos\theta \quad \text{(joules)}$$
where $\theta$ is the angle between the force and displacement.
Variable force (using integration):
$$W = \int_a^b F(x)\,dx$$
Example 1: A spring with spring constant $k = 200$ N/m is compressed 0.1 m. Work done:
$$W = \int_0^{0.1} 200x\,dx = [100x^2]_0^{0.1} = 1 \text{ J}$$
| Type | Formula | Notes |
|---|---|---|
| Kinetic | $KE = \frac{1}{2}mv^2$ | Depends on speed only |
| Gravitational PE | $GPE = mgh$ | Relative to reference level |
| Elastic PE | $EPE = \frac{1}{2}kx^2$ | Spring extension/compression |
Total mechanical energy: $E = KE + PE$.
$$W_{\text{net}} = \Delta KE = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$$
Proof: $W = \int F\,dx = \int ma\,dx = m\int v\,dv = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$. $\square$
When only conservative forces act (gravity, spring — not friction):
$$KE_1 + PE_1 = KE_2 + PE_2$$
Example 2: A 5 kg block slides down a smooth ramp of height 3 m, starting from rest. Find the speed at the bottom.
$$\frac{1}{2}mv^2 = mgh \Rightarrow v = \sqrt{2gh} = \sqrt{2\times9.8\times3} = \sqrt{58.8} \approx 7.67 \text{ m/s}$$
When friction does work:
$$KE_2 + PE_2 = KE_1 + PE_1 - W_{\text{friction}}$$
where $W_{\text{friction}} = \mu_k N \cdot d$ (always positive, representing energy removed from the mechanical system).
Example 3: A 4 kg block ($\mu_k = 0.3$) slides 5 m along a horizontal surface starting at 10 m/s.
$W_{\text{friction}} = 0.3\times(4\times9.8)\times5 = 58.8$ J.
$KE_2 = \frac{1}{2}(4)(100) - 58.8 = 200 - 58.8 = 141.2$ J.
$v_2 = \sqrt{2\times141.2/4} = \sqrt{70.6} \approx 8.40$ m/s.
$$P = \frac{dW}{dt} = \mathbf{F}\cdot\mathbf{v} = Fv\cos\theta \quad \text{(watts)}$$
Average power: $\bar{P} = W/t$.
Example 4: A car (mass 1200 kg) accelerates from 0 to 30 m/s in 10 s on a level road. Average power:
$\bar{P} = \Delta KE / t = \frac{1}{2}(1200)(900)/10 = 54000$ W $= 54$ kW.
KEY TAKEAWAY: The work-energy theorem directly links the net work done to the change in kinetic energy. Energy conservation eliminates the need to find acceleration when speed is sufficient.
EXAM TIP: Choose energy methods when the problem asks for speed without asking for time or acceleration. Choose Newton’s second law when time or forces are explicitly needed.
COMMON MISTAKE: Omitting the friction energy loss when applying conservation of energy. In real systems, $E_1 = E_2 + W_{\text{friction}}$, not $E_1 = E_2$.
