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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{-2}$ ]
Solution
Speed of conveyor belt, $v=2 \mathrm{~m} / \mathrm{s}$
<br/><br/>Coefficient of friction between the conveyor belt and bag, $\mu=0.4$
<br/><br/>Acceleration of bag due to slipping motion,
<br/><br/>$$
a=\mu g=0.4 \times 10=4 \mathrm{~m} / \mathrm{s}^2 \quad\left(\because g=10 \mathrm{~m} / \mathrm{s}^2\right)
$$
<br/><br/>If $s$ be the distance travelled by the bag on the belt during slipping motion, then
<br/><br/>$
\ v^2 \ =u^2-2 a s $<br/><br/>$
\Rightarrow \ 0=v^2-2 a s $<br/><br/>$
\Rightarrow s \ =\frac{v^2}{2 a}=\frac{2^2}{2 \times 4}=0.5 \mathrm{~m}
$
About this question
Subject: Physics · Chapter: Laws of Motion · Topic: Newton's First Law
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