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?

Question

Jarrett Pierce

Physics

18 December, 10:05

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?

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Answers (1)

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Joe Skinner · Answer 1

W = 4 mgh

Explanation:

Work is the product of distance by barefoot in this case

W = F. d = F d cos o

Where the angle between force and displacement, which is zero, cos 0º = 1

With the elevator going constant speed the force that pulls it is equal to its weight

F = mg

W = Fd d

The distance traveled is the height of each floor, the total height

d = 4 h

Therefore, the work is

W = mg 4h