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,

.....(i)

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