An elevator of mass m is initially at rest on the first floorof a building. It m

Question

An elevator of mass m is initially at rest on the first floorof a building. It moves upward, and passes the second andthrid floors with a constant velocity, and finally stops at thefourth floor. The distance between adjacent floors ish. What is the network done on the levator during the entiretrip, from the first floor to the fourth floor? a) W = -3 mgh b) None of these. c) W=-4 mgh d) W = 4 mgh e) W = 3 mgh f) W = 0 An elevator of mass m is initially at rest on the first floorof a building. It moves upward, and passes the second andthrid floors with a constant velocity, and finally stops at thefourth floor. The distance between adjacent floors ish. What is the network done on the levator during the entiretrip, from the first floor to the fourth floor? a) W = -3 mgh b) None of these. c) W=-4 mgh d) W = 4 mgh e) W = 3 mgh f) W = 0

Explanation / Answer

Taking the bottom of the building as the reference for thepotential energy . Initially , Potential energy = 0 Kinetic energy = 0 (As initial velocity is zero ) Finally , Kinetic energy = 0 (As final velocity is zero ) Potential energy = mg(height ) = 3mgh From work energy theorem , Net work done = change in energy Here change in energy = final - initial energy = 3 mgh - 0 =3mgh So Net work done = change in energy = 3mgh

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