1.

The conveyed belt is moving at `4m//s` the coefficient of static friction between the conveyed belt and the 10kg package Bis ` mu_(s)=0.2 determine the shortest time in which the belt can be stopped so the the package does not slide on the belt A. 1sB. 2sC. 4sD. 8s

Answer» Correct Answer - B
`a_(max)=mu,g=0.2xx10=2m//s^(2)`
`t_(min)=u/a_(max)=4/2=2sec`


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