A 4000 kg elevator rises from the basement to the top of a100-m tall building. Which of the following is closest to the network done on the elevator as it rises through the middle 20m of itstrip at a constant velocity? Use g=10m/s^2.

A.) 800,000 J

B.) - 800,000J

C.) 80,000 J

D.) - 80,000J

E.) 0 J

E) 0 J

Explanation:

Hi there!

The net work done on the elevator can be calculated by adding the work done by the forces acting on the elevator. In this case, we can distinguish two forces acting in the vertical direction, the force that elevates the elevator (in the upward direction) and the gravity force (in the downward direction). Using Newton's second law in the vertical direction:

F - Fg = m · a

Where:

F = force of the elevator.

Fg = force of gravity.

m = mass of the elevator.

a = acceleration.

Since the elevator is moving at constant velocity, the acceleration is zero. Then:

F - Fg = 0

F = Fg

Now, let's calculate the work done by these forces during 20 m. The equation of work is the following:

W = F · d

Where:

W = work.

F = applied force.

d = distance.

The net work done on the elevator can be expressed as the work done by the elevator plus the work done by gravity:

Wnet = W elevator + W gravity

Notice that gravity does work in the direction opposite to the movement of the elevator, then, it has to be negative:

Wnet = W elevator - W gravity

Wnet = F · d - Fg · d

Wnet = (F - Fg) · d

Since F - Fg = 0

Then:

Wnet = (F - Fg) · d = 0 N · d = 0 J

The correct answer is E) 0 J.