A conveyor belt is moving at a constant speed of 2 ms-1. 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 ms-2, is

Question

A conveyor belt is moving at a constant speed of 2 ms-1. 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 ms-2, is (A) 0.4 m (B) 1.2 m (C) 0.6 m (D) Zero

Tardigrade Admin · Accepted Answer

Force of friction, f=μ m g ∴ a=(f/m)=(μ m g/m) =μ g=0.5 × 10=5 m s -2 Using v2-u2=2 a S 02-22=2(-5) × S ⇒ S=0.4 m

Q. 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 $μ = 0.5$ . The distance that the box will move relative to belt before coming to rest on it, taking $g = 10 m s^{- 2}$ , is

0.4 m

1.2 m

0.6 m

Zero

Force of friction, $f = μ m g$
$∴ a = \frac{f}{m} = \frac{μ m g}{m}$
$= μg = 0.5 \times 10 = 5 m s^{- 2}$
Using $v^{2} - u^{2} = 2 a S$
$0^{2} - 2^{2} = 2 (- 5) \times S$
$\Rightarrow S = 0.4 m$

Q. A conveyor belt is moving at a constant speed of 2ms−1. 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=10ms−2, is