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

Amelia Adams · Accepted Answer

The coefficient of static friction is #=0.12#

The coefficient of static friction is

#mu_s=F_r/N#

#F_r=k*x#

The spring constant is #k=4kgs^-2#

The compression is #x=3/2m#

Therefore,

#F_r=4*3/2=6N#

The normal force is

#N=mg=5gN#

The coefficient of static friction is

#F_r=(6)/(5g)=0.12#

Lily Adams · Answer

undefined To find the minimum value of the surface's coefficient of static friction, use the equation $ f_{\text{friction}} = k \cdot x $. Here, $ k $ is the spring constant and $ x $ is the compression of the spring. Substituting the given values, the minimum value of the coefficient of static friction would be $ f_{\text{friction}} = 4 \cdot \frac{3}{2} $ = 6 N. Since the object is just about to move, the friction force equals the maximum static friction force, $ f_{\text{friction}} = \mu_s \cdot N $, where $ \mu_s $ is the coefficient of static friction and $ N $ is the normal force. Thus, $ \mu_s \cdot m \cdot g = 6 $, where $ m $ is the mass of the object and $ g $ is the acceleration due to gravity. Solving for $ \mu_s $, we get $ \mu_s = \frac{6}{m \cdot g} = \frac{6}{5 \cdot 9.8} \approx 0.122 $. Therefore, the minimum value of the surface's coefficient of static friction is approximately 0.122.