Question:
In a factory it is desired to lift $2000 \mathrm{~kg}$ of metal through a distance of $12 \mathrm{~m}$ in 1 minute. Find the minimum horsepower of the engine to be used.
Solution:
$W=F d \cos \theta$
$W=m g \cos \theta=2000 \times 10 \times 12 \times 1$
$W=24 \times 10^{4} J$
$P=\stackrel{\frac{w}{t}}{P}=\left(24 \times 10^{4}\right) / 60=4000 \mathrm{~W}$
$P=\frac{W}{t}=\left(24 \times 10^{4}\right) / 60=4000 W$
Horse power $=\mathrm{P} / 746=4000 / 746=5.3 \mathrm{hp}$