An object with a mass of # 5 kg# is lying still on a surface and is compressing a horizontal spring by #2 m#. If the spring's constant is # 6 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

1 Answer
Jul 13, 2018

Answer:

# mu_s = 0.245#

Explanation:

This problem works easier if the spring constant is in units of #N/m#. So I will prove that #kg/s^2 -= N/m#.

Multiply both sides of #kg/s^2 = N/m# by m,

#m * kg/s^2 = cancel(m) * N/cancel(m)#

Now we see that the expression is

#kg*m/s^2 = N#

These are the units of Newton's 2nd Law. So truly, #kg/s^2 -= N/m# and our spring constant can be written #6 N/m#.

Since the spring has been compressed by 2 m, the force it is exerting on the object is

#6 N/cancel(m) * 2 cancel(m) = 12 N#

Since the object remains "still", the static friction, #F_s#, between the object and the surface must be 12 N -- in the opposite direction. The formula that gives the relationship between the force of static friction, #F_s#; the coefficient of static friction, #mu_s#; and the normal force pressing the 2 surfaces together, #N# is

#F_s = mu_s*N#

And #N# in this case is the weight, #W#, of the object, given by

#W = m*g = 5 kg*9.8 m/s^2 = 49 N#

Going back to the formula for #F_s#, and remembering that the static friction, #F_s#, between the object and the surface must be 12 N,

#12 N = mu_s*49 N#

Solving for #mu_s#

# mu_s = (12 N) / (49 N) = 0.245#

That is the minimum coefficient of static friction that will allow the object to remain still. A larger # mu_s# will also hold the object.

I hope this helps,
Steve