A conveyor belt is moving at a constant speed of 2 m s-1. A box is gently dropped on it. The coefficient of friction between them is = 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
- 0.4 m
- 1.2 m
- 0.6 m
- zero
Answer
(1) 0.4 m
Solution
In the frame of the belt, the box starts with speed 2 m s-1 (opposite to the belt's motion) and is retarded by friction.
Retardation, a = g = 0.5 10 = 5 m s-2
Using v2 = u2 - 2as with v = 0:
s = u22a = (2)22 5
= 0.4 m
Answer: (1)