Question

An object with a mass of `12 kg` is lying still on a surface and is compressing a horizontal spring by `1 m`. If the spring's constant is `2 (kg)/s^2`, what is the minimum value of the surface's coefficient of static friction?

Answer 1

`1/60` (assuming `g=10 m s^{-2}`)

Explanation:

The force that the spring applies on the body is given by `k Delta x = 2` N. This must be balanced by the force of static friction, and so the limiting friction must exceed this. Now the normal force exerted by the surface on the object must balance its weight and is thus `120 N` (taking the acceleration due to gravity to be `10 m s^{-2}` ). Thus

`mu times 120 >= 2 implies mu >= 1/60`