A motor with an efficiency of 75% lifts a 120 kg load at a steady 1.25 m/s…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A motor with an efficiency of $75\%$ lifts a $120\ \text{kg}$ load at a steady $1.25\ \text{m/s}$ . Taking $g = 9.8\ \text{m/s}^2$ , find the useful power and the power the motor takes from the supply.

Need a reminder? The key ideaShow

Power is the rate of doing work: $P = \dfrac{W}{t}$ , in watts. For a force at speed $v$ , $P = Fv$ . Efficiency $=$ useful power out $\div$ power in. One kilowatt-hour is $3.6 \times 10^{6}\ \text{J}$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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