Question 94: Laws of Motion

Question

A bag is gently dropped on a conveyor belt moving at a speed of $$2 \mathrm{~m} / \mathrm{s}$$. The coefficient of friction between the conveyor belt and bag is $$0.4$$. 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 $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{-2}$$ ]

Accepted Answer

Speed of conveyor belt, $v=2 \mathrm{~m} / \mathrm{s}$
Coefficient of friction between the conveyor belt and bag, $\mu=0.4$
Acceleration of bag due to slipping motion,
$$
a=\mu g=0.4 	imes 10=4 \mathrm{~m} / \mathrm{s}^2 \quad\left(\because g=10 \mathrm{~m} / \mathrm{s}^2ight)
$$
If $s$ be the distance travelled by the bag on the belt during slipping motion, then
$
\ v^2 \ =u^2-2 a s $
$
\Rightarrow \ 0=v^2-2 a s $
$
\Rightarrow  s \ =\frac{v^2}{2 a}=\frac{2^2}{2 	imes 4}=0.5 \mathrm{~m}
$

Answer

2 m

Answer

0.5 m

Answer

3.2 m

Answer

0.8 ms

Laws of Motion

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