# An object with a mass of  18 kg is lying on a surface and is compressing a horizontal spring by 60 cm. If the spring's constant is  8 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?

Jan 25, 2016

$\setminus {\mu}_{s} = 0.027$

#### Explanation:

The friction force is dependent on 2 factors.
The coefficient of static friction($\setminus {\mu}_{s}$) and the force the body exerts on the surface.

In the above question, the body is attached to a compressed spring, and hence the spring exerts some force on the body. This force is given as $F = k x$. $k$ is given to be $8 k g \left({s}^{-} 2\right)$ and $x$ to be $0.6 m$ So force acting on the body is $F = 8 \cdot 0.6 = 4.8 N$

(Note that ${F}_{\text{ext}}$ here refers to the spring force)

Now, we know that the body is at rest because no force is acting on it, or the vectorial sum of the forces is zero. Which means that the spring force is being equalized by the force of friction.

Now we know that the force of friction is given by $f = \setminus {\mu}_{s} N$
where $N$ is the force that is exerted by the surface on the body and $\setminus {\mu}_{s}$ the coefficient of friction.
We know that the body is on a horizontal platform without angle, so $N = m g \setminus \implies f = \setminus {\mu}_{s} m g$

Since the force of friction and force of spring are equal (opposite in direction remind you), we get $k x = \setminus {\mu}_{s} m g$
Substituting the values, you'll get the above given answer.