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)`

Options

A. `5.0ms^(-2)`
B. `3.0ms^(-2)`
C. `1.5ms^(-2)`
D. `15ms^(-2)`

Answer

b

Explanation

:m=50kg
`mu=0.3`
`g=10ms^(-2)`
Suppose a is the common acceleration of the box and the train Block will be at rest if limiting friction `(f_(I)` is equal to ma (a=maximum acceleration).
`f_(1)maRightarrowmuN=ma`
`Rightarrowmumg=maRightarrowa=mug`
`therefore a=0.3xx10=3ms^(-2)`