A 400kg elevator starts from rest,exelerates uniformly to a constant speed of 2.0m/s and accelerates uniformly to a stop 20m above its friction and other losses,what work is done on the elevator?
To calculate the work done on the elevator, we need to find the total change in potential energy and the total change in kinetic energy.
Given data:
Initial velocity, u = 0 m/s
Final velocity, v = 2.0 m/s
Mass of the elevator, m = 400 kg
Height, h = 20 m
1. Change in potential energy:
ΔPE = mgh
ΔPE = 400 kg * 9.81 m/s^2 * 20 m
ΔPE = 78,480 J
2. Change in kinetic energy:
Initial kinetic energy, KE1 = 0
Final kinetic energy, KE2 = 0.5mv^2
KE2 = 0.5 * 400 kg * (2.0 m/s)^2
KE2 = 800 J
3. Total work done on the elevator:
Work = ΔPE + ΔKE
Work = 78,480 J + 800 J
Work = 79,280 J
Therefore, the work done on the elevator is 79,280 Joules.
Ah, there is no change in kinetic energy. It starts at a stop and stops at a stop.
You are correct. Since the elevator starts from rest and ends at rest, there is no change in kinetic energy. Therefore, the work done on the elevator is equal to the change in potential energy only.
Given data:
Mass of the elevator, m = 400 kg
Height, h = 20 m
Acceleration due to gravity, g = 9.81 m/s^2
Work done on the elevator = ΔPE = mgh
Work = 400 kg * 9.81 m/s^2 * 20 m
Work = 78,480 J
Therefore, the work done on the elevator is 78,480 Joules.