Work-Energy Theorem Physics Example 1

Solution

1. 1
   The work-energy theorem states: $W_{\text{net}} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.
2. 2
   Net work done: $W = Fd = 20 \times 5 = 100 \text{ J}$.
3. 3
   Since $v_i = 0$: $$100 = \frac{1}{2}(4)v_f^2 \implies v_f = \sqrt{\frac{200}{4}} = \sqrt{50} \approx 7.07 \text{ m/s}$$