Conveyor at 4 m/s, block enters at 8 m/s

A horizontal conveyor belt runs at a steady 4 m/s. A small block slides onto the belt at 8 m/s in the same direction as the belt. The coefficient of kinetic friction between block and belt is 0.2 (g = 10). 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 = |4−8|/2 = 2
   Friction erases the speed difference second by second
3. s = |v_belt² − v₀²|/(2a) = 12
   Displacement before the speeds match

Why it works

The block is faster than the belt, so the sliding friction acts backwards on it and slows it down steadily until it is moving at the belt's 4 m/s.

Common mistakes

• Making the block speed up because the belt keeps moving.
• Using the belt's 4 m/s as the block's speed for the whole slide.

Related

conveyor-belt meeting-problem conveyor-12ms-block-4ms