CalcElfCalcElf › Conveyor at 6 m/s, block enters at 2 m/s
Conveyor at 6 m/s, block enters at 2 m/s
A horizontal conveyor belt runs at a steady 6 m/s. A small block is placed on it already moving at 2 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.
Answer
Takes 2 s to reach a common speed, displacement 8 m
Step-by-step
- a = μg = 0.2×10 = 2Kinetic friction supplies the acceleration
- t = Δv/a = |6−2|/2 = 2Friction erases the speed difference second by second
- s = |v_belt² − v₀²|/(2a) = 8Displacement before the speeds match
Why it works
Because the block is slower than the belt, the friction from the belt points forward and speeds the block up at a constant rate until the two speeds match.
Common mistakes
- Pointing the friction backwards because the belt rubs the block.
- Using the belt's 6 m/s in the distance formula instead of the block's average speed.