An object with a mass of #12 kg# is lying still on a surface and is compressing a horizontal spring by #1/3 m#. If the spring's constant is #4 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

1 Answer
P dilip_k
Apr 23, 2016

Force F required to compress a spring by x unit is
#F=k*x#,where k = force constant.
Here #x =1/3m#
#k=4kgs^-2#
hence Force exerted on the object of mass 1 kg is #F=4(kg)/s^2xx1/3m=4/3(kg*m)/s^2=4/3N#
This force is balanced by the then static frictional force as it is self adjusting one,

So the minimum value of the coefficient of static frictions
#mu_s=F/N#,where N is the normal reaction exerted on the body by the floor. Here #N =mg=12*9.8N#
So
#mu_s=F/N=(4/3)/(9.8xx12)=0.011#

Related questions
See all questions in Frictional Forces
Impact of this question
1154 views around the world
You can reuse this answer
Creative Commons License