# An object with a mass of  5 kg is lying still on a surface and is compressing a horizontal spring by 60 c m. If the spring's constant is  6 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?

Dec 30, 2016

To balance the force due to the spring with the force of static friction requires a static coefficient of 0.022

#### Explanation:

Since the mass is stationary, we know that the forces acting on it must be in balance. These forces are the elastic force of the spring and the force of friction

${F}_{\text{spring}} = {F}_{f}$

$\frac{1}{2} k {x}^{2} = {\mu}_{s} {F}_{N}$

On a horizontal surface, the normal force is equal to the force of gravity acting on the mass.

$\frac{1}{2} k {x}^{2} = {\mu}_{s} m g$

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

${\mu}_{s} = \frac{k {x}^{2}}{2 m g}$

Now, inserting the numbers (and changing 60 cm into 0.6 m)

${\mu}_{s} = \frac{6 {\left(.6\right)}^{2}}{2 \left(5\right) 9.8} = 0.022$