An elevator motor provides 45.0 kW of power while lifting an elevator 35.0 m. If the elevator contains seven passengers each with an average mass of 70.0 kg and it takes 20.0 s to accomplish this task, determine the mass of the elevator.

Find how much work ∆W is done by the motor in lifting the elevator:

P = ∆W / ∆t

where

• P = 45.0 kW = power provided by the motor

• ∆W = work done

• ∆t = 20.0 s = duration of time

Solve for ∆W :

∆W = P ∆t = (45.0 kW) (20.0 s) = 900 kJ

In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is M = (m + 490.0) kg, where m is the mass of the elevator alone. Then

∆W = M g h

where

• g = 9.80 m/s² = acceleration due to gravity

• h = 35.0 m = distance covered by the elevator

Solve for M, then for m :

M = ∆W / (g h) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg

m = M - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg

i believe the answer is a

we need to know the time interval during which the force acts. impulse of force = momentum change