JEE Main Physics question

PhysicsLaws of MotionJEE Main

A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. Then the coefficient of friction is

Answer: (D)

Solution

Deceleration a = Δv/t = 6/10 = 0.6 m/s². μ = a/g = 0.6/10 = 0.06.

Question ID VK-153461