Verified Solution • 2025–26 Session moderate

QJEEM22Ph7914: JEE Main 2022 Physics - Class 11 JEEM22Ph7914 Solution: A bag is gently dropped on a conveyor belt moving at a speed of …

A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s2 \mathrm{~m} / \mathrm{s} . 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=10 m/s−2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{-2} ]

Multiple Choice Options

  • A 2 m
  • B 0.5 m
  • C 3.2 m
  • D 0.8 ms

Step-by-Step Verified Solution

Speed of conveyor belt, v=2 m/sv=2 \mathrm{~m} / \mathrm{s}

Coefficient of friction between the conveyor belt and bag, μ=0.4\mu=0.4

Acceleration of bag due to slipping motion,

a=μg=0.4×10=4 m/s2(∵g=10 m/s2)a=\mu g=0.4 \times 10=4 \mathrm{~m} / \mathrm{s}^2 \quad\left(\because g=10 \mathrm{~m} / \mathrm{s}^2\right)

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

 v2 =u2−2as\ v^2 \ =u^2-2 a s

⇒ 0=v2−2as\Rightarrow \ 0=v^2-2 a s

⇒s =v22a=222×4=0.5 m\Rightarrow s \ =\frac{v^2}{2 a}=\frac{2^2}{2 \times 4}=0.5 \mathrm{~m}

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