A body of mass 2 kg is placed on a horizontal surface having coefficient of kinetic friction 0.4 and limiting coefficient of static friction 0.5. If a horizontal force 2.5 N is applied on the body, the frictional force acting on the body will be (Take, `g = 10ms^(2)`)

Options

A. 8 N
B. 10 N
C. 20 N
D. 2.5 N

Answer

D

Explanation

`f_("static")=(mu_("static"))mg`
`implies f_("static")-(0.5)(2)(10)implies f_("static")-10 N`
`because` Applied force is just 2.5 N `lt` 10 N. So, the block will not move. As, we know that in static region, friction is a self adjusting force and has a value equal to that of applied force.
`"So, "f=F_("applied")=2.5 N`