Saved Bookmarks
| 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` |
|