Power input

The construction lift lifted a 300 kg load to a height of 12 m with uniform movement. If the efficiency is 75% and the engine has a power input of 5 kW, how long did it take for the load to be lifted?

Correct answer:

t =  9.6 s

Step-by-step explanation:

$$\begin{gathered} P=5\text{kW}\to \text{W}=5\cdot 1000\text{W}=5000\text{W} \\ \xi =75\%=\frac{75}{100}=\frac{3}{4}=0.75 \\ V=\xi \cdot P=0.75\cdot 5000=3750\text{W} \\ m=300\text{kg} \\ h=12\text{m} \\ g=10{\text{m/s}}^2 \\ F=m\cdot g=300\cdot 10=3000\text{N} \\ W=F\cdot h=3000\cdot 12=36000\text{J} \\ t=W/V=36000/3750=9.6\text{s} \end{gathered}$$

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