CalcElfCalcElf › Conveyor at 5 m/s, block starts from rest
Conveyor at 5 m/s, block starts from rest
A horizontal conveyor belt runs at a steady 5 m/s. A small block is set down gently on the belt with zero initial speed. The coefficient of kinetic friction between block and belt is 0.5 (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 1 s to reach a common speed, displacement 2.5 m
Step-by-step
- a = μg = 0.5×10 = 5Kinetic friction supplies the acceleration
- t = Δv/a = |5−0|/5 = 1Friction erases the speed difference second by second
- s = |v_belt² − v₀²|/(2a) = 2.5Displacement before the speeds match
Why it works
While the block is slower than the belt, sliding friction pulls it forward with a constant acceleration, and it stops accelerating the moment the two speeds match.
Common mistakes
- Using the belt speed as the block's speed from the very first instant.
- Forgetting that the friction is kinetic here, so friction equals μ times the weight.