A 1000 kg elevator rises from rest in the basement to the fourth floor,a distance of 20m.As it passes the fourth floor its speed is 4m/s. There is a constant frictional force of 500N. The work done by the lifting mechanism is:

(A)196*10^3 J

(B)204*10^3 J

(C)214*10^3 J

(D)203*10^5 J

Let F be the upward force acting on the lift.

Friction force (f) and the weight (mg) of the lift acting the downward direction,

Consider that lift is moving up with an acceleration a

so, according to the problem,


s = 20 m

F is the force by lifting machine in upward direction.

Work done = F*s      ....(ii)

On substituting and solving we get,

Work done = 214*10^3 J

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