Recall and apply the principle of conservation of energy to a variety of physical situations.
The total energy of an isolated system remains constant; it can be transferred between forms or between objects, but it cannot be created or destroyed.
Mathematically, for an isolated system:
$$\Delta E_{\text{total}} = 0$$or equivalently
$$E_{\text{initial}} = E_{\text{final}}$$The work done on an object equals the change in its kinetic energy:
$$W_{\text{net}} = \Delta K = K_{\text{final}} - K_{\text{initial}}$$When only conservative forces act, the work done can be expressed as the negative change in potential energy, leading to the mechanical energy conservation equation:
$$K_{\text{i}} + U_{\text{i}} = K_{\text{f}} + U_{\text{f}}$$Follow these steps when solving a problem:
A 2.0 kg ball is dropped from a height of 5.0 m. Neglect air resistance. Find its speed just before it hits the ground.
Solution:
$$\begin{aligned} &\text{Initial energy: }E_i = K_i + U_i = 0 + mgh = (2.0)(9.8)(5.0) = 98\ \text{J}\\ &\text{Final energy: }E_f = K_f + U_f = \frac{1}{2}mv^2 + 0\\ &\text{Conservation: }E_i = E_f \\ &\frac{1}{2}mv^2 = 98\ \text{J} \\ &v = \sqrt{\frac{2\times98}{2.0}} = \sqrt{98} \approx 9.9\ \text{m s}^{-1} \end{aligned}$$A cart of mass 0.5 kg is attached to a horizontal spring (constant $k = 200\ \text{N m}^{-1}$) compressed by 0.15 m. The cart is released on a frictionless track. Determine the speed of the cart when the spring returns to its natural length.
Solution:
$$\begin{aligned} &U_{\text{spring}} = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.15)^2 = 2.25\ \text{J}\\ &K_{\text{final}} = \frac{1}{2}mv^2 = U_{\text{spring}}\\ &\frac{1}{2}(0.5)v^2 = 2.25\\ &v^2 = \frac{2\times2.25}{0.5}=9\\ &v = 3.0\ \text{m s}^{-1} \end{aligned}$$| Question | Key Concepts | Answer (for teacher) |
|---|---|---|
| A 1.5 kg block slides down a frictionless 30° incline from a height of 2.0 m. Find its speed at the bottom. | Gravitational PE → KE, $mgh = \frac12 mv^2$ | $v = \sqrt{2gh} = \sqrt{2\times9.8\times2.0} \approx 6.3\ \text{m s}^{-1}$ |
| A 0.8 kg pendulum bob is released from a horizontal position. What is its speed at the lowest point? (Neglect air resistance.) | PE loss = KE gain, $mgh = \frac12 mv^2$ with $h = L(1-\cos\theta)$ | $v = \sqrt{2gL}$ (for $\theta = 90^\circ$, $h = L$) |
| A 3.0 kg cart moving at 4.0 m s⁻¹ encounters a rough patch that does 12 J of negative work. What is its speed after the patch? | Work–energy theorem, $W_{\text{nc}} = \Delta K$ | $\frac12 m v_f^2 = \frac12 m v_i^2 + W_{\text{nc}}$ → $v_f = \sqrt{v_i^2 + 2W_{\text{nc}}/m} \approx 2.9\ \text{m s}^{-1}$ |