example\na 1000-kg elevator carries a maximum load of 800 kg. a constant frictional force of 4000 n retards…

example\na 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?

example\na 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?

Answer

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)).