An object with a mass of # 3 # #kg# is lying on a surface and is compressing a horizontal spring by #10 # #cm#. If the spring's constant is # 9# #kgs^-2#, what is the minimum value of the surface's coefficient of static friction?

1 Answer
Apr 7, 2016

Answer:

The frictional coefficient #mu=0.03#.

Explanation:

(convert the distance to SI units: #10# #cm# = #0.1# #m#)

The force exerted by the spring is given by:

#F=kx=9xx0.1=0.9# #N#

Since the object is stationary, this force applied by the spring must be being balanced by the frictional force.

The frictional force is given by:

#F_"frict"=muF_"normal"#

And the normal force is the gravitational force:

#F_"normal"=mg=3xx9.8=29.4# #N#

Rearranging:

#mu=F_"frict"/F_"normal"=0.9/29.4=0.03#

(frictional coefficients do not have units)

(side note: I keep saying that #kgs^-2# is an odd unit for spring constant. It is correct, but so is #Nm^-1#, and the latter makes is easier to understand what is going on in most contexts.)