An elevator can carry a maximum load of $1800 \; kg$ (elevator $+$ passengers) and is moving up with a constant speed of $2 \; m s^{-1}$. The frictional force opposing the motion is $4000 \; N$. Determine the minimum power delivered by the motor to the elevator in watts as well as in horsepower.

(N/A) The total downward force acting on the elevator is the sum of the gravitational force and the frictional force.
$F = mg + F_{f} = (1800 \times 10) + 4000 = 18000 + 4000 = 22000 \; N$.
Since the elevator moves at a constant speed,the motor must exert an upward force equal to the total downward force.
Power $P$ is given by the product of force and velocity: $P = F \cdot v$.
$P = 22000 \; N \times 2 \; m s^{-1} = 44000 \; W$.
To convert power into horsepower $(hp)$,we use the conversion factor $1 \; hp = 746 \; W$.
$P = \frac{44000}{746} \approx 58.98 \; hp \approx 59 \; hp$.