Question

An object with a mass of `5 kg` is lying still on a surface and is compressing a horizontal spring by `2 m`. If the spring's constant is `8 (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 =2m
`k=8kgs^-2`
hence Force exerted on the object of mass 1 kg is `F=8(kg)/s^2*2m=16(kg*m)/s^2=16N`
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=5*9.8N`
So
`mu_s=F/N=16/(9.8xx5)=0..33`