Conveyor at 8 m/s, block enters at 20 m/s

A horizontal conveyor belt runs at a steady 8 m/s. A small block slides onto the belt at 20 m/s in the same direction as the belt. The coefficient of kinetic friction between block and belt is 0.4 (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 = 0.4×10 = 4
   Kinetic friction supplies the acceleration
2. t = Δv/a = |8−20|/4 = 3
   Friction erases the speed difference second by second
3. s = |v_belt² − v₀²|/(2a) = 42
   Displacement before the speeds match

Why it works

The block moves faster than the belt, so friction points backwards and slows the block at a constant a = μg = 4 m/s² until its speed drops to the belt's 8 m/s.

Common mistakes

• Multiplying the belt's 8 m/s by the time instead of using the block's average speed of 14 m/s.
• Expecting the block to reverse and be carried backwards by the belt.

Related

conveyor-belt incline-friction conveyor-6ms-block-2ms