# 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?

Jul 13, 2018

${\mu}_{s} = 0.245$

#### Explanation:

This problem works easier if the spring constant is in units of $\frac{N}{m}$. So I will prove that $k \frac{g}{s} ^ 2 \equiv \frac{N}{m}$.

Multiply both sides of $k \frac{g}{s} ^ 2 = \frac{N}{m}$ by m,

$m \cdot k \frac{g}{s} ^ 2 = \cancel{m} \cdot \frac{N}{\cancel{m}}$

Now we see that the expression is

$k g \cdot \frac{m}{s} ^ 2 = N$

These are the units of Newton's 2nd Law. So truly, $k \frac{g}{s} ^ 2 \equiv \frac{N}{m}$ and our spring constant can be written $6 \frac{N}{m}$.

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

$6 \frac{N}{\cancel{m}} \cdot 2 \cancel{m} = 12 N$

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

${F}_{s} = {\mu}_{s} \cdot N$

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

$W = m \cdot g = 5 k g \cdot 9.8 \frac{m}{s} ^ 2 = 49 N$

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,

$12 N = {\mu}_{s} \cdot 49 N$

Solving for ${\mu}_{s}$

${\mu}_{s} = \frac{12 N}{49 N} = 0.245$

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