Question

An electric hoist is used to lift a 235.0 kg load to a height of 69.0 m in 38.1 s. (a) what is the power of the hoist in kw?

Answer 1

$Given:\\m=235.0kg\\h=69.0m\\t=38.1s\\g=9.81 \frac{m}{s^2} \\\\Find:\\P=?\\\\Solution:\\\\P= \frac{W}{t}\\\\W=\Delta E_p\\\\E_p_0=0\Rightarrow \Delta E_p=E_p\\\\E_p=mgh \\\\P= \frac{mgh}{t} \\\\P= \frac{235kg\cdot9.81 \frac{m}{s^2}\cdot69m}{38.1s} \approx 4175W=4.175kW$