A suitcase is gently placed on a horizontal conveyor belt. It leaves a skid mark of length 2.5 m on the belt before it stops slipping and moves along with the belt. If the coefficient of kinetic friction between the suitcase and the belt is 0.5 , the speed of the conveyor belt is: [Take g = 10 m s ⁻² ]

Step-by-step solution

The acceleration of the suitcase due to kinetic friction is a = g = 0.5 10 = 5 m s ⁻² . The length of the skid mark represents the relative distance s_ rel slipped by the suitcase on the belt. The initial relative velocity of the belt with respect to the suitcase is u_ rel = v (where v is the speed of the belt). The final relative velocity when slipping stops is v_ rel = 0 . Using the kinematic equation for relative motion: v_ rel ^2 = u_ rel ^2 - 2 a s_ rel 0 = v^2 - 2(5)(2.5) v^2 = 25 v = 5 m s ⁻¹ . Answer: 5 m s