A machine of efficiency of 70% is used to raise a body of mass 80 kg through a vertical distance of 3 m in 40 seconds. Calculate the power input (Take g=10 m/s2).

The work done by the machine is to raise the body, which is equal to the change in its potential energy.

$$W_{out} = mgh$$

Substitute the given values:

$$W_{out} = (80 \text{ kg})(10 \text{ m/s}^2)(3 \text{ m})$$ $$W_{out} = 2400 \text{ J}$$

Power output is the work output divided by the time taken.

$$P_{out} = \frac{W_{out}}{t}$$

Substitute the calculated work and given time:

$$P_{out} = \frac{2400 \text{ J}}{40 \text{ s}}$$ $$P_{out} = 60 \text{ W}$$

The efficiency of a machine is the ratio of power output to power input.

$$\eta = \frac{P_{out}}{P_{in}}$$

Rearrange the formula to solve for $P_{in}$:

$$P_{in} = \frac{P_{out}}{\eta}$$

Substitute the calculated power output and given efficiency:

$$P_{in} = \frac{60 \text{ W}}{0.70}$$ $$P_{in} \approx 85.71 \text{ W}$$

The power input is $85.71 \text{ W}$.