A 100 kg block is started with a speed of 2.0 m s ⁻¹ on a long, rough belt maintained in a fixed horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20 . Consider the scenario from a frame of reference travelling at 2.0 m s ⁻¹ parallel to the initial velocity of the block. As seen from this frame, the block is gently placed on a moving belt, and in due time it begins moving wi

Step-by-step solution

Let the ground frame be S and the moving frame be S' . In the ground frame S , the initial velocity of the block is v_ i = 2.0 m s ⁻¹ and the belt is stationary ( v_ belt = 0 ). The frame S' is moving with a velocity v_ F = 2.0 m s ⁻¹ parallel to the block's initial velocity. Initial velocity of the block in frame S' is: u' = v_ i - v_ F = 2.0 - 2.0 = 0 m s ⁻¹ When the block ceases slipping, it moves together with the stationary belt. The final velocity of the block in the ground frame S is v_ f = 0 m s ⁻¹ . Final