An object with a mass of # 14 kg# is lying still on a surface and is compressing a horizontal spring by #1 m#. If the spring's constant is # 7 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

1 Answer
Jan 18, 2018

Answer:

The surface's coefficient of static friction, in this case, will be 0.0051, approximately.

Explanation:

Since the object is lying still on the surface, it is in equilibrium and, therefore, the net force in it must equal zero:

#F_N = 0#. (1)

Since it is placed by a horizontal spring, the friction force, #f#, will also be in that direction, opposing to the spring's force.

Now, since the displacement #Deltax = 1m# and #k = 7 N/m#, then the elastic force is

#F = k . Deltax#;

#F = 7 . 1 = 7 N#.

Now, from Equation (1), #f# and #F# must have equal intensities, which means that

#f = 7 N#.

Now, since #f = mu . N#, with #mu# and #N# being the surface's coefficient of static friction and #N# the object's normal force, respectively, then:

#mu . N = 7#.

Since in this case the normal force equals the object's weight, then:

#mu = 7/(9,81 . 14)#

#mu = 0.051#, approximately (I have used #g = 9.81 m/s^2).