Question

If 250 J of work is done in sliding a 5 kg block up an inclined plane of height 4 m. work done against friction is ( $ g = 10m/{s^2} $ )
(A) 50 J
(B) 100 J
(C) 200 J
(D) zero

Answer 1

Hint : We need to take the total work done to be equal to the work done against gravity plus the work done against friction. The work done against gravity is simply the product of the weight of the body and the height moved by the block.

Formula used: In this solution we will be using the following formula;
 $ w = mg $ where $ w $ is the weight of a body, $ m $ is the mass of the body, and $ g $ is the acceleration due to gravity.
 $ {W_g} = wh $ where $ {W_g} $ is the work done against gravity, and $ w $ is the weight of the body, and $ h $ is the height moved by the body.

Complete step by step answer:
We asked to calculate the work done against friction given the total work done, the mass of the body and height moved by the body.
In this case, it is safe to assume that the total work done is the work done against friction plus the work done against gravity.
Hence,
 $ {W_t} = {W_g} + {W_f} $ where $ {W_t} $ is total work, $ {W_g} $ is the work done against gravity, and $ {W_f} $ is the work done against friction.
But $ {W_g} = wh $ where $ w $ is the weight of the body, and $ h $ is the height moved by the body. And
 $ w = mg $ where $ m $ is the mass of the body, and $ g $ is the acceleration due to gravity.
Hence,
 $ {W_f} = {W_t} - mgh $
From the question,
 $ {W_f} = 250 - 5\left( {10} \right)\left( 4 \right) = 250 - 200 = 50{\text{J}} $
Hence, the correct option is A.

Note:
 For clarity, it is not a fundamental law or something that the total work done is equal to the work done by friction and gravity. The situation only works when the body is moving up at a constant velocity.
Generally, the work done by the force is the value of the force itself times the length of the inclined plane. However, if the force is equal to the weight component in the direction parallel to the inclined plane [and parallel (actually anti parallel) to the force] plus the friction against the motion, then the work done by the force will indeed be equal to the sum of the work done by friction and gravity.