Q. A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is μ = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g = 10 m s⁻², is
A) Zero
B) 0.4 m
C) 1.2 m
D) 0.6 m
Explanation
[Solution from source] Relative to the belt the box starts at 2 m/s and friction gives it a relative deceleration μg = 5 m/s². Relative distance = u²/2a = 4/10 = 0.4 m. The answer zero ignores that the box slips until it reaches the belt speed. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.