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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. None of these.
3. W = 0
4. W = 4 m g h
5. W = ?3 m g h
6. W = 3 m g h

Respuesta :

Answer:

6. W = 3 m g h

Explanation:

Displacement in moving from first floor to the fourth floor is 3h, and the total energy consumed in this process is (3.m.g.h) which is equal to the work done.

According to the work-energy equivalence the change in energy of a system is equal to the work done.

Here, change in potential energy:

[tex]\Delta PE=m.g(\Delta h)[/tex]

[tex]\Delta PE=3mgh[/tex]

[tex]\therefore W=\Delta PE[/tex]

[tex]W=3mgh[/tex]

Answer:

W = 0

Explanation:

The formula for Work is Force multiplied by distance. For there to be a force, there must be acceleration. In this case, the elevator has a constant velocity, and so the acceleration is 0. If the acceleration is zero, the force is zero, and so the work is zero.

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