JEE Main Physics question
300 joule of work is done in sliding up a 2 kg block on an inclined plane to a height of 10 metres. Taking value of acceleration due to gravity 'g' to be 10 m/s², work done against friction is
Answer: (A)
Solution
Work done against gravity = mgh = 2×10×10 = 200 J. Work done against friction = Total work − work against gravity = 300 − 200 = 100 J.