Question

An object with a mass of `10 kg` is lying still on a surface and is compressing a horizontal spring by `5/6 m`. If the spring's constant is `1 (kg)/s^2`, what is the minimum value of the surface's coefficient of static friction?

Answer 1

`0.008`, rounded to three decimal places.

Explanation:

Equation for the spring is
`F=-kx`

where `k` is spring constant and `x` is the elongation or compression of the spring. `-ve` sign indicates that restoring force acts opposite to the deformation of the spring.

In the problem spring's force in equilibrium with the mass compressing the spring is
`F=-1xx5/6=-5/6N`
Force due to static friction `F_s=mu_sxx"Normal Reaction"`.

We also know that
`"Normal Reaction"="mass of the object" xx g` , `g` is gravity and `=9.81ms^-2`

or `F_s=mu_sxx10xx9.81`.
Equating the magnitudes of two forces in equilibrium
`mu_sxx10xx9.81=5/6`, solving for minimum coefficient of static friction
`mu_s=5/6xx1/98.1=0.008`, rounded to three decimal places.