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. None of these.
4. W = 3 m g h
5. W = 4 m g h
6. W = −3 m g h
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Comments
Kate
30.11.20, 00:36
Actually, the answer is 0. The elevator starts and ends at a state of
rest which means KEi and KEf = 0 because 1/2m(0)^2 = 0. If KEi and KEf
= 0, then the change in KE = 0, and since the change in KE = W, W = 0.
Hope that helps.
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Comments
Actually, the answer is 0. The elevator starts and ends at a state of rest which means KEi and KEf = 0 because 1/2m(0)^2 = 0. If KEi and KEf = 0, then the change in KE = 0, and since the change in KE = W, W = 0. Hope that helps.
Leave a comment