An object with a mass of #12 kg# is lying still on a surface and is compressing a horizontal spring by #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
Mar 19, 2016

#mu =36/(12xx10)=0.3#

Explanation:

We know that the gravitational Force applied or normal reaction #F_N= m*g=12*10 N# [ Taking g= 10 m/s^2]
force of limiting value of static friction F= #mu* F_N=mu*12*10N#
Again #F =1/2kx^2= 1/2*4*3^2=36N#, where k = force constant = #4(kg)/s^2# x = amount of compression of spring '
So
#mu*12*10=36#
#mu =36/(12xx10)=0.3#