Question

example
a 1000-kg elevator carries a maximum load of 800 kg. a constant frictional force of 4000 n retards its motion upward. what minimum power must the motor deliver to lift the fully loaded elevator at a constant speed of 3 m/s?

Explanation:

Step1: Calculate total mass

The elevator mass is \(1000\space kg\) and load is \(800\space kg\), so total mass \(m = 1000 + 800 = 1800\space kg\).

Step2: Calculate gravitational force

Gravitational force \(F_g = mg\), where \(g = 9.8\space m/s^2\). So \(F_g = 1800\times9.8 = 17640\space N\).

Step3: Determine total force to overcome

Frictional force \(F_f = 4000\space N\). Since the elevator moves at constant speed (acceleration \(a = 0\)), the tension \(T\) (force from motor) must balance \(F_g\) and \(F_f\). So \(T = F_g + F_f = 17640 + 4000 = 21640\space N\).

Step4: Calculate power

Power \(P = Fv\) (since \(F\) and \(v\) are in same direction), \(v = 3\space m/s\). So \(P = 21640\times3 = 64920\space W = 64.92\space kW\).

Answer:

The minimum power the motor must deliver is \(64920\space W\) (or \(64.92\space kW\)).