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 `6 (kg)/s^2`, what is the minimum value of the surface's coefficient of static friction?

Answer 1

`mu_s = 0.245`

Explanation:

This problem works easier if the spring constant is in units of `N/m`. So I will prove that `kg/s^2 -= N/m`.

Multiply both sides of `kg/s^2 = N/m` by m,

`m * kg/s^2 = cancel(m) * N/cancel(m)`

Now we see that the expression is

`kg*m/s^2 = N`

These are the units of Newton's 2nd Law. So truly, `kg/s^2 -= N/m` and our spring constant can be written `6 N/m`.

Since the spring has been compressed by 2 m, the force it is exerting on the object is

`6 N/cancel(m) * 2 cancel(m) = 12 N`

Since the object remains "still", the static friction, `F_s`, between the object and the surface must be 12 N -- in the opposite direction. The formula that gives the relationship between the force of static friction, `F_s`; the coefficient of static friction, `mu_s`; and the normal force pressing the 2 surfaces together, `N` is

`F_s = mu_s*N`

And `N` in this case is the weight, `W`, of the object, given by

`W = m*g = 5 kg*9.8 m/s^2 = 49 N`

Going back to the formula for `F_s`, and remembering that the static friction, `F_s`, between the object and the surface must be 12 N,

`12 N = mu_s*49 N`

Solving for `mu_s`

`mu_s = (12 N) / (49 N) = 0.245`

That is the minimum coefficient of static friction that will allow the object to remain still. A larger `mu_s` will also hold the object.

I hope this helps,
Steve