Conveyor at 12 m/s, block enters at 4 m/s

A horizontal conveyor belt runs at a steady 12 m/s. A small block slides onto the belt at 4 m/s in the same direction as the belt. The coefficient of kinetic friction between block and belt is 0.2 (g = 10 m/s²). Find the time needed for the block to reach the belt's speed, and the distance the block slides relative to the ground in that time.

Step-by-step

1. a = μg = 0.2×10 = 2
   Kinetic friction supplies the acceleration
2. t = Δv/a = |12−4|/2 = 4
   Friction erases the speed difference second by second
3. s = |v_belt² − v₀²|/(2a) = 32
   Displacement before the speeds match

Why it works

Because the block is slower than the belt, friction from the belt acts forward and gives the block the constant acceleration μg until its speed equals the belt's.

Common mistakes

• Assuming the block instantly takes the belt's 12 m/s.
• Using a = g = 10 m/s² and forgetting to multiply by the friction coefficient.

Related

conveyor-belt incline-friction conveyor-10ms-block-4ms