An object with a mass of # 12 kg# is lying on a surface and is compressing a horizontal spring by #40 cm#. 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
Jul 14, 2017

Answer:

The coefficient of static friction is #=0.014#

Explanation:

The coefficient of static friction is

#mu_s=F_r/N#

#F_r=k*x#

The spring constant is #k=4kgs^-2#

The compression is #x=0.4m#

Therefore,

#F_r=4*0.4=1.6N#

The normal force is

#N=mg=12gN#

The coefficient of static friction is

#mu_s=(1.6)/(12g)=0.014#