1.

maximum acceleration of the train in which a 50 Kg bx lying on its florr will remain stationary (Given : Co - efficient f static friction between the box and the trains floor is 0.3 and g=10ms^(-2)

Answer»

`5.0ms^(-2)`
`3.0ms^(-2)`
`1.5ms^(-2)`
`15ms^(-2)`

Solution :FRICTION accelerates the box.
`F=ma`
But `F_(MAX)=MUMG`
So `F_(max)=ma_(max)`
`mumg=ma_(max)`
`a_(max)=mug=0.3xx10=3m//s^(2)`


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