An inclined plane has a load of 800 N and an effort of 350 N. The vertical height is 1.5 m and the distance along the plane is 4.0 m. Calculate the efficiency and the work done against friction.

Work output is the useful work done, which is lifting the load vertically.

$$W_{out} = L \times h$$ $$W_{out} = 800 \, \text{N} \times 1.5 \, \text{m}$$ $$W_{out} = 1200 \, \text{J}$$

Work input is the total work done by the effort along the inclined plane.

$$W_{in} = E \times d$$ $$W_{in} = 350 \, \text{N} \times 4.0 \, \text{m}$$ $$W_{in} = 1400 \, \text{J}$$

Efficiency is the ratio of work output to work input, expressed as a percentage.

$$\eta = \frac{W_{out}}{W_{in}} \times 100\%$$ $$\eta = \frac{1200 \, \text{J}}{1400 \, \text{J}} \times 100\%$$ $$\eta = \frac{12}{14} \times 100\%$$ $$\eta = \frac{6}{7} \times 100\%$$ $$\eta \approx 0.8571 \times 100\%$$ $$\eta \approx \boxed{\textbf{85.7\%}}$$

2. Find the work done against friction as the load is raised.

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

$$W_f = W_{in} - W_{out}$$ $$W_f = 1400 \, \text{J} - 1200 \, \text{J}$$ $$W_f = \boxed{\textbf{200 \, J}}$$