Question

An object with a mass of `12 kg` is lying still on a surface and is compressing a horizontal spring by `1/3 m`. If the spring's constant is `4 (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 =1/3m`
`k=4kgs^-2`
hence Force exerted on the object of mass 1 kg is `F=4(kg)/s^2xx1/3m=4/3(kg*m)/s^2=4/3N`
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=12*9.8N`
So
`mu_s=F/N=(4/3)/(9.8xx12)=0.011`