Question

An elevator of massmis initially at rest onthe first floor of a building. It moves upward,and passes the second and third floors witha constant velocity, and finally stops at thefourth floor.The distance between adjacentfloors is h.What is the net work done on the elevatorduring the entire trip, from the first floor tothe fourth floor?
1. None of these.
2. W=-3m g h
3. W=-4m g h
4. W= 3m g h
5. W= 0correct
6. W= 4m g h

Answer 1

To solve this problem it is necessary to apply the concepts of Work. Work is understood as the force applied to travel a determined distance, in this case the height. The force in turn can be expressed by Newton's second law as the ratio between mass and gravity, as well

$W = mgh$

Where,

m = mass

h = height

g = Gravitational constant

When it ascends to the second floor it has traveled the energy necessary to climb a height, under this logic, until the 4 floor has traveled 3 times the height h of each of the floors therefore

$h = 3h$

Replacing in our equation we have to

$W = mgh\\W = mg(3h)\\W = 3mgh$

The correct answer is 4.