A block-and-tackle system is used to lift a load of 20 N through a vertical height of 10 m. if the efficiency of the system is 40%, how much work is done against friction?

The useful work done is the work done on the load.

$$W_{\text{out}} = \text{Load} \times \text{Vertical height}$$ $$W_{\text{out}} = 20 \text{ N} \times 10 \text{ m}$$ $$W_{\text{out}} = 200 \text{ J}$$

Efficiency ($\eta$) is given by:

$$\eta = \frac{W_{\text{out}}}{W_{\text{in}}} \times 100\%$$

Given $\eta = 40\%$, we can rearrange to find $W_{\text{in}}$:

$$W_{\text{in}} = \frac{W_{\text{out}}}{\eta}$$ $$W_{\text{in}} = \frac{200 \text{ J}}{0.40}$$ $$W_{\text{in}} = 500 \text{ J}$$

The work done against friction is the difference between the total work input and the useful work output.

$$W_{\text{friction}} = W_{\text{in}} - W_{\text{out}}$$ $$W_{\text{friction}} = 500 \text{ J} - 200 \text{ J}$$ $$W_{\text{friction}} = 300 \text{ J}$$

The work done against friction is $\text{300 J}$.