Conveyor at 5 m/s, block enters at 15 m/s

A horizontal conveyor belt runs at a steady 5 m/s. A small block slides onto the belt at 15 m/s in the same direction as the belt. The coefficient of kinetic friction between block and belt is 0.5 (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 = 0.5×10 = 5
   Kinetic friction supplies the acceleration
2. t = Δv/a = |5−15|/5 = 2
   Friction erases the speed difference second by second
3. s = |v_belt² − v₀²|/(2a) = 20
   Displacement before the speeds match

Why it works

The block slides forward relative to the belt, so the friction on it points backwards and gives a steady deceleration a = μg = 5 m/s² until the speeds are equal.

Common mistakes

• Taking the coefficient 0.5 itself as the acceleration, which gives 0.5 m/s².
• Believing the block keeps slowing down and stops after reaching the belt's speed.

Related

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