# An object with a mass of 4 kg is lying still on a surface and is compressing a horizontal spring by 7/8 m. If the spring's constant is 12 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?

Dec 16, 2017

The coefficient of static friction must be at least 0.268

#### Explanation:

If the object is motionless, then Newton's first law assures us that the (two) horizontal forces acting on it must be is balance.

These forces are the elastic force of the spring, which is described by Hooke's law:

$F = k x$ where $k$ is the spring constant and x is the amount of compression (or extension) of the spring.

and the force of friction:

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

where ${F}_{N}$ is the normal force between the mass and the surface it rests on, and in the case of a horizontal surface, is equal to $m g$.

Putting it together, we get

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

and, solving for ${\mu}_{s}$:

${\mu}_{s} = \frac{k x}{m g} = \frac{12 \times \frac{7}{8}}{4 \times 9.8} = \frac{10.5}{39.2} = 0.268$

Dec 16, 2017

The coefficient of static friction is $= 0.27$

#### Explanation:

The mass is $m = 4 k g$

The compression of the spring is $x = \frac{7}{8} m$

The spring constant is $k = 12 k g {s}^{-} 2$

The reaction of the spring is $R = k x = 12 \cdot \frac{7}{8} = 10.5 N$

THe acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

The normal reaction of the object is $N = m g = 4 \cdot 9.8 = 39.2 N$

The coefficient of static friction is

${\mu}_{s} = \frac{R}{N} = \frac{10.5}{39.2} = 0.27$