An object with a mass of  14 kg is lying still on a surface and is compressing a horizontal spring by 1 m. If the spring's constant is  7 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?

Jan 18, 2018

Answer:

The surface's coefficient of static friction, in this case, will be 0.0051, approximately.

Explanation:

Since the object is lying still on the surface, it is in equilibrium and, therefore, the net force in it must equal zero:

${F}_{N} = 0$. (1)

Since it is placed by a horizontal spring, the friction force, $f$, will also be in that direction, opposing to the spring's force.

Now, since the displacement $\Delta x = 1 m$ and $k = 7 \frac{N}{m}$, then the elastic force is

$F = k . \Delta x$;

$F = 7 . 1 = 7 N$.

Now, from Equation (1), $f$ and $F$ must have equal intensities, which means that

$f = 7 N$.

Now, since $f = \mu . N$, with $\mu$ and $N$ being the surface's coefficient of static friction and $N$ the object's normal force, respectively, then:

$\mu . N = 7$.

Since in this case the normal force equals the object's weight, then:

$\mu = \frac{7}{9 , 81 . 14}$

$\mu = 0.051$, approximately (I have used #g = 9.81 m/s^2).