Question

A 1000 kg elevator rises from rest in the basement to the fourth floor, a distance of 20 m. As it passes the fourth floor its speed is 4 m s–1. There is a constant frictional force of 500 N. The work done by the lifting mechanism is
(Take g = 9.8 m s–2)

Answer 1

Work done= change in potential energy

=mgh=1000×9.8×20=196×10^3

Answer 2

Weight of the elevator = 1000kg (Given)

Distance = 20m (Given)

Speed = 4m/s (Given)

Let the upward force acting on the lift be = F

Let the lift move up with acceleration = a

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

Therefore,  

a = v² - u²/2s

According to second law of motion -  

>F - ( mg + f) = ma -- eq 1

Friction force (f) and the weight (mg) of the lift acts towards the upwards direction,

Work done = F s -- eq 2  

Thus, work done = 214 × 10 ×3 J