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?

1 Answer
A08
Apr 26, 2016

#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.

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