An object with a mass of 15 kg is lying still on a surface and is compressing a horizontal spring by 5/6 m. If the spring's constant is 15 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?

Mar 30, 2016

$\mu = 0 , 085$

Explanation:

$F = {u}_{k} \cdot \Delta x$

${u}_{k} = 15 \text{ } \frac{k g}{s} ^ 2$

$\Delta x = \frac{5}{6} \text{ } m$

$F = 15 \cdot \frac{5}{6} = \frac{75}{6} = \frac{25}{2} \text{ N}$

${F}_{f} = \mu \cdot N \text{( Friction Force)}$
$\mu : \text{coefficient of static friction}$
$N : \text{normal force to the contacting surfaces}$
$\text{we can write "N=m*g" (The Newton's third law)}$

$F = {F}_{f}$
$\text{m:15 kg}$
$\frac{25}{2} = \mu \cdot 15 \cdot 9 , 81$

$25 = 2 \cdot 15 \cdot 9 , 81 \cdot \mu$

$25 = 294 , 3 \cdot \mu$

$\mu = \frac{25}{294 , 3}$

$\mu = 0 , 085$