A machine lifts 30 tonnes of weed to a height of 15 m in 5 minutes. The input power of the machine is 20 kW. Calculate the efficiency of the machine.

The mass of weed ($m$) is 30 tonnes.

$$m = 30 \text{ tonnes} = 30 \times 1000 \text{ kg} = 30000 \text{ kg}$$

The input power ($P_{in}$) of the machine is 20 kW.

$$P_{in} = 20 \text{ kW} = 20 \times 1000 \text{ W} = 20000 \text{ W}$$

The time ($t$) taken is 5 minutes.

$$t = 5 \text{ minutes} = 5 \times 60 \text{ s} = 300 \text{ s}$$

The height ($h$) the weed is lifted is 15 m.

$$h = 15 \text{ m}$$

We use the acceleration due to gravity ($g$) as $9.81 \text{ m/s}^2$.

The useful work done is the potential energy gained by lifting the weed.

$$W = mgh$$ $$W = 30000 \text{ kg} \times 9.81 \text{ m/s}^2 \times 15 \text{ m}$$ $$W = 4414500 \text{ J}$$

Output power is the useful work done per unit time.

$$P_{out} = \frac{W}{t}$$ $$P_{out} = \frac{4414500 \text{ J}}{300 \text{ s}}$$ $$P_{out} = 14715 \text{ W}$$

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

$$\eta = \frac{P_{out}}{P_{in}} \times 100\%$$ $$\eta = \frac{14715 \text{ W}}{20000 \text{ W}} \times 100\%$$ $$\eta = 0.73575 \times 100\%$$ $$\eta = 73.575\%$$

Rounding to one decimal place:

$$\eta \approx 73.6\%$$

The efficiency of the machine is \text{73.6%}.