Conveyor at 6 m/s, block enters at 10 m/s

A horizontal conveyor belt runs at a steady 6 m/s. A small block slides onto the belt at 10 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 = |6−10|/2 = 2
   Friction erases the speed difference second by second
3. s = |v_belt² − v₀²|/(2a) = 16
   Displacement before the speeds match

Why it works

The block is faster than the belt, so sliding friction acts backwards on it and slows it down steadily until the block is moving at the belt's 6 m/s.

Common mistakes

• Thinking a faster block keeps running ahead and never slows to the belt speed.
• Using (10 − 6) ÷ 10 = 0.4 s, forgetting that the acceleration is μg = 2 m/s².

Related

conveyor-belt meeting-problem conveyor-5ms-block-at-rest