20. What is the net work done on the elevator during the entire trip, from the first floor to the fourth floor?

Question

An elevator of mass m is initially at rest on the first floor of a building. It moves upward, and passes the second and third floors with a constant velocity, and finally stops at the fourth floor. The distance between adjacent floors is h. An elevator of mass m is initially at rest on the first floor of a building. It moves upward, and passes the second and third floors with a constant velocity, and finally stops at the fourth floor. The distance between adjacent floors is h. What is the net work done on the elevator during the entire trip, from the first floor to the fourth floor?

1. W = −4 m g h

2. W = 0

3. W = 3 m g h

4. W = −3 m g h

5. None of these.

6. W = 4 m g h

Answer 1

Work done by a constant force is given by

W = (F cos θ) s

where F is the magnitude of the force,

s is the magnitude of the displacement,

and θ is the angle between the force and the displacement.

Here

F = mg since F_net = F – mg = 0 at constant velocity

s = 3h

and θ = 0° since force and displacement are in the same direction

So

W = (mg cos 0°)3h = 3mgh

Answer

3. W = 3mgh