JEE Main Physics question
A marble block of mass 2 kg lying on ice, when given a velocity of 6 ms⁻¹, is stopped by friction in 10 s. Then the coefficient of friction is
Answer: (C)
Solution
a = Δv/t = 6/10 = 0.6 ms⁻² = μg ⟹ μ = 0.6/10 = 0.06.
A marble block of mass 2 kg lying on ice, when given a velocity of 6 ms⁻¹, is stopped by friction in 10 s. Then the coefficient of friction is
a = Δv/t = 6/10 = 0.6 ms⁻² = μg ⟹ μ = 0.6/10 = 0.06.