An object with a mass of #16 kg# is lying still on a surface and is compressing a horizontal spring by #7/8 m#. If the spring's constant is #15 (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 #7/8   m#. If the spring's constant is #15    (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

Sebastian Acosta · Accepted Answer

The coeficient of friction is #=0.084#

The static friction coefficient is

#mu_s=F_r/N#

#F_r=k*x#

The spring constant is #k=15kgs^-2#

The compression is #x=7/8m#

Consequently,

#F_r=15*7/8=105/8N#

The typical force is

#N=mg=16gN#

The static friction coefficient is

#F_r=(105/8)/(16g)=13.125/(16g)=0.084#

Parker Abbott · Answer

undefined To find the minimum value of the coefficient of static friction, we need to consider the forces acting on the object when it is at rest. The force exerted by the spring, which is $ kx $, where $ k $ is the spring constant and $ x $ is the compression of the spring. The force due to gravity, $ mg $, where $ m $ is the mass of the object and $ g $ is the acceleration due to gravity. The maximum force of static friction is $ \mu_s N $, where $ \mu_s $ is the coefficient of static friction and $ N $ is the normal force. At the minimum value of $ \mu_s $, the static friction force balances the net force acting on the object. Therefore, we have:

$$ kx + \mu_s N = mg $$

$$ kx + \mu_s mg = mg $$

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

Substituting the given values:

$$ \mu_s = \frac{(15\, \text{kg}/\text{s}^2) \cdot (7/8\, \text{m})}{(16\, \text{kg}) \cdot (9.8\, \text{m/s}^2)} $$

$$ \mu_s \approx \frac{131.25}{156.8} $$

$$ \mu_s \approx 0.8366 $$

Therefore, the minimum value of the surface's coefficient of static friction is approximately $ \mu_s = 0.8366 $.