A block of mass 5kg is resting on a rough surface as shown in the figure. It is acted upon by a force of F towards right. Find frictional force acting on block when 50N `( mu_(s) =0.6 , mu_(k) = 0.5) [ g = 10ms^(-2) ]`

Explanation

Maximum value of frictional force that the surface can offer is
`f_("max") = f_("lim") = mu_(s) N `
`= 0.6 xx 5 xx 10 = 30 ` newton

Therefore, it `F le f_("max")` body will be at rest and f = F or `F gt f_("max")` body will more and `f= f_(k)`
`F =50 N gt F_("max")`
So body will move and kinetic frictional force will act, its value will be
`f_(k) = mu_(k) N = 0.5 xx 5 xx 10 = 25 ` newton