Efficiency Physics Example 2

Solution

1. 1
   Useful work output: $W_{\text{out}} = mgh = 200 \times 10 \times 10 = 20{,}000 \text{ J}$.
2. 2
   Using efficiency: $\eta = \frac{W_{\text{out}}}{W_{\text{in}}}$, so $W_{\text{in}} = \frac{W_{\text{out}}}{\eta}$.
3. 3
   $$W_{\text{in}} = \frac{20{,}000}{0.85} \approx 23{,}529 \text{ J}$$