JEE Main 2022 — Laws Of Motion Question with Solution

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,

$$a=\mu g=0.4\times 10=4\mathrm{m}/{\mathrm{s}}^2\left(\because g=10\mathrm{m}/{\mathrm{s}}^2\right)$$

If $s$ be the distance travelled by the bag on the belt during slipping motion, then

$ \ 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} $