FrictionHard
Question
A conveyor bet is moving at a constant speed of 2 ms-1. A box is gently dropped on it. The coefficient of friction between them them is μ = 0.5. the distance that the box will move relative to belt before coming to rest on it, taking g = 10 ms-2 , is
Options
A.0.4 m
B.1.2 m
C.0.6 m
D.zero
Solution
Force of friction, f = μ mg
∴
= μg = 0.5 × 10 = 5 ms-1
Using v2 - u2 = 2as
02 - 22 = 2(- 5)× S ⇒ S = 0.4 m
∴
= μg = 0.5 × 10 = 5 ms-1Using v2 - u2 = 2as
02 - 22 = 2(- 5)× S ⇒ S = 0.4 m
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