Object on a moving belt: time to match speed

A belt runs at 2 m/s. An object is placed on it at 0.5 m/s, friction coefficient 0.2 (g = 10). Find the time and distance until speeds match.

Step-by-step

1. a = μg = 0.2×10 = 2
   Kinetic friction supplies the acceleration
2. t = Δv/a = |2−0.5|/2 = 0.75
   Friction erases the speed difference second by second
3. s = |v_belt² − v₀²|/(2a) = 0.9375
   Displacement before the speeds match

Why it works

Sliding friction gives a constant acceleration a = μg. The object keeps accelerating until its speed equals the belt speed; the time comes from Δv/a and the distance from (v² − v₀²)/(2a).

Common mistakes

• Using Δv²/(2a) for the distance instead of (v² − v₀²)/(2a).
• Forgetting that the object stops accelerating once speeds match.

Related

incline-friction meeting-problem conveyor-8ms-block-20ms