CalcElf › Conveyor at 10 m/s, block enters at 4 m/s
Conveyor at 10 m/s, block enters at 4 m/s
A horizontal conveyor belt runs at a steady 10 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.
Answer
Takes 3 s to reach a common speed, displacement 21 m
Step-by-step
- a = μg = 0.2×10 = 2Kinetic friction supplies the acceleration
- t = Δv/a = |10−4|/2 = 3Friction erases the speed difference second by second
- s = |v_belt² − v₀²|/(2a) = 21Displacement before the speeds match
Why it works
The block is slower than the belt, so the sliding friction points forward and speeds the block up at a constant rate until the two speeds match.
Common mistakes
- Pointing the friction backwards and making the block slow down.
- Using the belt's 10 m/s for the distance instead of the block's average speed.