Question

An object with a mass of `14 kg` is lying still on a surface and is compressing a horizontal spring by `1 m`. If the spring's constant is `7 (kg)/s^2`, what is the minimum value of the surface's coefficient of static friction?

Answer 1

The surface's coefficient of static friction, in this case, will be 0.0051, approximately.

Explanation:

Since the object is lying still on the surface, it is in equilibrium and, therefore, the net force in it must equal zero:

`F_N = 0`. (1)

Since it is placed by a horizontal spring, the friction force, `f`, will also be in that direction, opposing to the spring's force.

Now, since the displacement `Deltax = 1m` and `k = 7 N/m`, then the elastic force is

`F = k . Deltax`;

`F = 7 . 1 = 7 N`.

Now, from Equation (1), `f` and `F` must have equal intensities, which means that

`f = 7 N`.

Now, since `f = mu . N`, with `mu` and `N` being the surface's coefficient of static friction and `N` the object's normal force, respectively, then:

`mu . N = 7`.

Since in this case the normal force equals the object's weight, then:

`mu = 7/(9,81 . 14)`

`mu = 0.051`, approximately (I have used #g = 9.81 m/s^2).