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Question: If 250 J of work is done in sliding a 5 kg block up an inclined plane of height 4 m. work done again...

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/s2g = 10m/{s^2} )
(A) 50 J
(B) 100 J
(C) 200 J
(D) zero

Explanation

Solution

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=mgw = mg where ww is the weight of a body, mm is the mass of the body, and gg is the acceleration due to gravity.
Wg=wh{W_g} = wh where Wg{W_g} is the work done against gravity, and ww is the weight of the body, and hh 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,
Wt=Wg+Wf{W_t} = {W_g} + {W_f} where Wt{W_t} is total work, Wg{W_g} is the work done against gravity, and Wf{W_f} is the work done against friction.
But Wg=wh{W_g} = wh where ww is the weight of the body, and hh is the height moved by the body. And
w=mgw = mg where mm is the mass of the body, and gg is the acceleration due to gravity.
Hence,
Wf=Wtmgh{W_f} = {W_t} - mgh
From the question,
Wf=2505(10)(4)=250200=50J{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.