A block of mass 1kg is at rest on a rough horizontal surface having coefficient of static friction 0.2 and kinetic friciton 0.15, find the frictional forces if a horizontal force,
(a) `F=1N` (b) F=1.96N (c) F=2.5N is applied on a block

Explanation

Maximum force of friction `f_("max")=0.2xx1xx9.8N=1.96N`
(a) For `F_("ext")=1N, F_("ext")ltf_("max")`
So, body is in rest means static friction is present and hence `f_(s)=F_("ext")=1N`
(b) For `F_("ext")=1.96N,F_("ext")=f_("max")=1.96N` So `f=1.96N`
(c) For `F_("ext")=2.5N`. So `F_("ext")gtf_("max")`
Now body is in moving condition `therefore f_("max")=f_(k)=mu_(k)N=mu_(k) mg=0.15xx1xx9.8=1.47N`