Laws of Motion

JEE Main level, single correct, medium

150 of 210

Question 150 of 210

A bag is gently dropped on a conveyor belt moving at a speed of 2ms−12m{s}^{-1}. The coefficient of friction between the conveyor belt and bag is 0.40.4 Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is : [Take g=10ms−2g=10m{s}^{-2}]

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Answer: (B) 0.5m0.5m

When the bag is dropped on the conveyor belt, the acceleration of the bag will be a=μg=4ms−2a=\mu g=4m{s}^{-2}
The bag will stop moving on the conveyor when the velocity of the bag and the velocity of the conveyor becomes equal.
u=0,v=2ms−1u=0,v=2m{s}^{-1}
Using the equation of motion for constant acceleration,
v2=u2+2as{v}^{2}={u}^{2}+2as
⇒s=0.5m\Rightarrow s=0.5m

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