JEE Main Physics question
A block of mass 10 kg slides down a rough slope which is inclined at 45° to the horizontal. The coefficient of sliding friction is 0.30. When the block has slid 5 m, the work done on the block by the force of friction is nearly
Answer: (B)
Solution
Friction force f = μmg cosθ = 0.30×10×10×cos45° ≈ 21.2 N. Work done by friction = −f×s = −21.2×5 ≈ −106 J = −75√2 J.