A bag is gently dropped on a conveyor belt moving at a speed of 2 m/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 g = 10m/s-2]

Option 1 - <p>2m</p>
Option 2 - <p>0.5m</p>
Option 3 - <p>3.2m</p>
Option 4 - <p>0.8ms</p>
12 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
A
7 months ago
Correct Option - 2
Detailed Solution:

  a = μ g = 0 . 4 * 1 0 = 4 m / s 2

Slipping stops when Vbag = Vconveyar

-> 0 + at = 2

-> t =    2 a = 2 4 = . 5 s e c

So, x =    1 2 a t 2 = 1 2 * 4 * 1 4 = 0 . 5 m

 

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