Question

An object with a mass of `8 kg` is lying still on a surface and is compressing a horizontal spring by `5/6 m`. If the spring's constant is `12 (kg)/s^2`, what is the minimum value of the surface's coefficient of static friction?

Answer 1

Force F required to compress a spring by x unit is
`F=k*x`,where k = force constant.
Here `x =5/6m`
`k=12kgs^-2`
hence Force exerted on the object of mass 1 kg is `F=12(kg)/s^2xx5/6m=10(kg*m)/s^2=10N`
This force is balanced by the then static frictional force as it is self adjusting one,

So the minimum value of the coefficient of static frictions
`mu_s=F/N`,where N is the normal reaction exerted on the body by the floor. Here `N =mg=8*9.8N`
So
`mu_s=F/N=10/(9.8xx8)=0.13`