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 ms -2, is

Question

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 ms -2, is (A) 1.2 m (B) 0.6 m (C) Zero (D) 0.4 m

Tardigrade Admin · Accepted Answer

The frictional force on the box f=μ m g ∴ Acceleration in the box a=μ g=5 m s-2 v2=u2+2 a s 4=0+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

$1.2 m$

$0.6 m$

Zero

$0.4 m$

The frictional force on the box $f = μ m g$
$∴$ Acceleration in the box
$a = μg = 5 m s^{- 2}$
$v^{2} = u^{2} + 2 a s$
$4 = 0 + 2 \times 5 \times s$
$s = 0.4 m$

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

1.2m

0.6m