An object with a mass of # 16 kg# is lying still on a surface and is compressing a horizontal spring by #5/3 m#. If the spring's constant is # 4 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

Question

An object with a mass of # 16  kg# is lying still on a surface and is compressing a horizontal spring by #5/3 m#. If the spring's constant is # 4 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

Elena Albarran · Accepted Answer

#mu=0.0425# Looking at the problem we can write the equilibrium force equation in horizontal direction: #"Friction force = Spring force"# So, #"Friction Coefficient"(mu) * "Reaction force" (R) = "Spring constant" (k) * "Displacement" (x)# #mu * R = k*x# ;or #mu * mg=k*x# ;or #mu * 16*9.81=4*5/3# ;or #mu=0.0425# Thus the Friction coefficient is #mu=0.0425#

Nathan Ashcroft · Answer

undefined The minimum value of the surface's coefficient of static friction can be found using the equation:

$$ \mu_s = \frac{kx}{mg} $$

where:
- $ \mu_s $ is the coefficient of static friction,
- $ k $ is the spring constant (4 kg/s²),
- $ x $ is the compression of the spring (5/3 m),
- $ m $ is the mass of the object (16 kg), and
- $ g $ is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the given values into the equation:

$$ \mu_s = \frac{(4 \, \text{kg/s}^2) \times (5/3 \, \text{m})}{(16 \, \text{kg}) \times (9.8 \, \text{m/s}^2)} $$

$$ \mu_s = \frac{20/3}{156.8} $$

$$ \mu_s \approx 0.042 $$

So, the minimum value of the surface's coefficient of static friction is approximately $ \mu_s \approx 0.042 $.