A conveyor belt is moving horizontally at a speed of 4 ms-1. A box of mass 20 kg is gently laid on it. It takes 0.1 s for the box to come to rest. If the belt continues to move uniformly, then the distance moved by the box on the conveyor belt is

Question

A conveyor belt is moving horizontally at a speed of 4 ms-1. A box of mass 20 kg is gently laid on it. It takes 0.1 s for the box to come to rest. If the belt continues to move uniformly, then the distance moved by the box on the conveyor belt is (A) zero (B) 0.2 m (C) 0.4 m (D) 0.8m

Tardigrade Admin · Accepted Answer

From first equation of motion v = u + at 0 = -4 + a × (0.1) ⇒ a=40 4ms-2 s=(v2/2a) ⇒ s=((4)2/2 × 40) ⇒ s=(16/80) ⇒ s=0.2 m

Q. A conveyor belt is moving horizontally at a speed of $4 m s^{- 1}$ . A box of mass 20 kg is gently laid on it. It takes 0.1 s for the box to come to rest. If the belt continues to move uniformly, then the distance moved by the box on the conveyor belt is

zero

0.2 m

0.4 m

0.8m

From first equation of motion
v = u + at
0 = -4 + a $\times (0.1)$
$\Rightarrow$ a=40 $4 m s^{- 2}$
$s = \frac{v ^{2}}{2 a}$
$\Rightarrow s = \frac{( 4 ) ^{2}}{2 \times 40}$
$\Rightarrow s = \frac{16}{80} \Rightarrow s = 0.2 m$

Q. A conveyor belt is moving horizontally at a speed of 4ms−1. A box of mass 20 kg is gently laid on it. It takes 0.1 s for the box to come to rest. If the belt continues to move uniformly, then the distance moved by the box on the conveyor belt is