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  • A bag is gently dropped on a conveyor belt moving at a speed of  <img alt="\mathrm{2 \mathrm{~m} / \mathrm{s}. }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B2%20%5Cm

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

Option: 1

\mathrm{2 \mathrm{~m}}


Option: 2

\mathrm{0.5 \mathrm{~m}}


Option: 3

\mathrm{3.2 \mathrm{~m}}


Option: 4

\mathrm{0.8 \mathrm{~ms}}


Answers (1)

best_answer

\mathrm{f=\mu m g}

\mathrm{a=\frac{f}{m}=\mu g}

Bag corne to rest when the speed of the bag is same as the conveyor belt ie.,\mathrm{ 4 \mathrm{~m} / \mathrm{s}}

\mathrm{ v^2=u^2+2 \text { as } }

\mathrm{ (4)^2=0+2(0.4 \times 10) s }

\mathrm{ S=2 \mathrm{~m} }
The distance travelled by the bag on the belt during slipping motion is \mathrm{2 \mathrm{~m} }

Hence 1 is correct option.

Posted by

Suraj Bhandari

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