# An object with a mass of  12 kg is lying on a surface and is compressing a horizontal spring by 50 cm. 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, 2016

${\mu}_{s} \ge 0.025$

#### Explanation:

The situation here is that the spring is already compressed by $0.5 m$ and the object is at rest as the surface is with friction.

That means force of friction and spring force are balancing each other out.

${F}_{s} = {F}_{f}$

Spring forces is ${F}_{s} = k x$ Given $k = 6 k \frac{g}{s} ^ 2$ and $x = 0.5 m$
${F}_{s} = 6 \cdot 0.5 = 3 N$

Now, frictional force depends on the normal force of the body. Since the surface is horizontal, the entire weight of the body is the normal force. So, ${F}_{f} = \setminus {\mu}_{s} \cdot N = \setminus {\mu}_{s} \cdot m \cdot g = \setminus {\mu}_{s} \cdot 12 \cdot 10$

Equating the two, ${\cancel{3}}^{1} = \setminus {\mu}_{s} \cdot {\cancel{120}}^{40}$

So, that's how the minimum static constant is so small.