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?

Question

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?

Charlotte Aaron · Accepted Answer

# mu_s = 0.245#

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

Sebastian Acosta · Answer

undefined To find the minimum value of the coefficient of static friction, we need to consider the forces acting on the object. When the spring is compressed, it exerts a force given by Hooke's Law, F_spring = k * x, where k is the spring constant and x is the displacement from the equilibrium position. Thus, F_spring = 6 (kg/s^2) * 2 m = 12 N. To keep the object in place, the static friction force must counteract the force exerted by the spring. Therefore, the minimum value of static friction force is equal to the force exerted by the spring, which is 12 N. Since static friction force is given by F_friction = μ_s * N, where μ_s is the coefficient of static friction and N is the normal force, we can rearrange the equation to find μ_s = F_friction / N. The normal force, N, is equal to the weight of the object, which is mg = 5 kg * 9.8 m/s^2 = 49 N. Therefore, μ_s = 12 N / 49 N ≈ 0.245. Hence, the minimum value of the surface's coefficient of static friction is approximately 0.245.