An object with a mass of # 12 kg# is lying still on a surface and is compressing a horizontal spring by #3/2 m#. If the spring's constant is # 2 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2017-04-27T05:44:26

The coefficient of static friction is #=0.026#

The force of friction is

#F_r=k*Deltax#

The spring constant is #k=2kgs^-2#

The distance is #Deltax=3/2m#

Therefore,

#F_r=2*3/2=3N#

The coefficient of static friction is

#mu_s=F_r/N#

The normal force is #N=mg=(12g) N#

So,

#mu_s=(3)/(12g)=0.026#