JEE Main Physics question
A block of mass 1 kg slides down on a rough inclined plane of inclination 60° starting from its top. If the coefficient of kinetic friction is 0.5 and length of the plane is 1 m, then work done against friction is (Take g = 9.8 m/s²)
Answer: (C)
Solution
Work done against friction = μmg cosθ × L = 0.5×1×9.8×cos60°×1 = 0.5×9.8×0.5 = 2.45 J.