Question

A horizontal force is applied on a body on a rough horizontal surface produces an acceleration`a`. If coefficient of friction between the body and surface which is `u` is reduced to `u /3`, the acceleration increases by 2 units. What is the value of `u`?

Answer 1

The value is `mu=3/g`

Explanation:

The force applied to body `=F`

The acceleration is `=a`

According to Newton's Second Law,

`F=ma`

Let the mass of the body be `=mkg`

The normal force is `N=mg N`

The force of friction is `=F_r`

The coefficient of friction is

`mu=F_r/N=F_r/(mg)`

`F_r=mumg`

`F-F_r=ma`

`F-mumg=ma`

`F=ma+mumg`

And

`F-mu/3mg=m(a+2)`

`F=m(a+2)+mu/3mg`

`ma+mumg=m(a+2)+mu/3mg`

`a+mug=a+2+mu/3g`

`2/3mug=2`

`mu=3/g`