Question

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/sec. there is a constant frictional force of 500N. The work done by lifting mechanism is:
(A) $ 196 \times {10^3}{\text{J}} $
(B) $ 204 \times {10^3}{\text{J}} $
(C) $ 214 \times {10^3}{\text{J}} $
(D) $ 203 \times {10^5}{\text{J}} $

Answer 1

Hint: Since the object is accelerating, then work done is additionally done in accelerating the elevator. The total work done is the sum of the work done to overcome friction, plus the work done to overcome gravity and the work done to accelerate the elevator.

Formula used: In this solution we will be using the following formulae;
 $ W = Fd $ where $ W $ is the work done on a body, $ F $ is the force acting on the body, and $ d $ is the distance moved by the object.
 $ {v^2} = {u^2} + 2as $ where $ v $ is the final velocity of a body, $ u $ is the initial, $ a $ is the acceleration of the body, and $ s $ is the distance moved by the body.

Complete Step-by-Step solution:
To solve the question, we note that the work done on the elevator is equal to the work done to overcome friction, plus the work done to overcome gravity and the work done to accelerate the elevator.
Generally, the work done by a force (or to overcome a force) can be given as
 $ W = Fd $ where $ W $ is the work done on a body, $ F $ is the force acting on the body, and $ d $ is the distance moved by the object.
Hence, work done to overcome friction is
 $ W = 500 \times 20 = 10000{\text{J}} $
To overcome gravity, is given as
 $ W = mgh = 1000 \times 9.8 \times 20 = 196000{\text{J}} $
And finally, the work done to accelerate the object would be given as
 $ W = mas $
We need to find the acceleration of the elevator.
From the third equation of motion, we have,
 $ {v^2} = {u^2} + 2as $ $ v $ is the final velocity of a body, $ u $ is the initial, $ a $ is the acceleration of the body, and $ s $ is the distance moved by the body.
The elevator started from rest, hence.
 $ {v^2} = 2as $ , observe that, if we divide both sides by 2 and multiply by $ m $ , we have
 $ m\dfrac{{{v^2}}}{2} = mas $
Hence,
 $ W = m\dfrac{{{v^2}}}{2} $
So, we have
 $ W = 1000\dfrac{{{4^2}}}{2}\left( {20} \right) = 160000{\text{J}} $
Now we add together, as in
 $ W = 10000{\text{J}} + 196000{\text{J}} + 160000{\text{J}} = 214000J $
 $ \Rightarrow W = 214 \times {10^3}J $
The correct option is C.

Note:
For clarity, the work done overcoming gravity is the work done which will just allow the elevator to start moving upward at a constant velocity. But since the object is accelerating, it means that the force applied must be greater than the force needed to overcome the weight.