An object with a mass of # 3 kg# is lying on a surface and is compressing a horizontal spring by #10 cm#. If the spring's constant is # 8 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

The coefficient of static friction is

Therefore,

The normal force is

The coefficient of static friction is

To find the minimum value of the surface's coefficient of static friction, we need to consider the forces acting on the object. The force exerted by the spring can be calculated using Hooke's Law, which states that the force exerted by a spring is proportional to its displacement from equilibrium. The force exerted by the spring ( F_s ) can be calculated using the formula ( F_s = k \cdot x ), where ( k ) is the spring constant and ( x ) is the displacement from equilibrium. In this case, ( k = 8 , \text{(kg/s}^2) ) and ( x = 10 , \text{cm} = 0.1 , \text{m} ).

So, ( F_s = 8 , \text{(kg/s}^2) \times 0.1 , \text{m} = 0.8 , \text{N} ).

Since the object is in equilibrium, the force exerted by the spring must be balanced by the force of static friction acting in the opposite direction. The maximum force of static friction ( F_{\text{max}} ) can be calculated using the formula ( F_{\text{max}} = \mu_s \cdot N ), where ( \mu_s ) is the coefficient of static friction and ( N ) is the normal force.

The normal force ( N ) can be calculated using the formula ( N = mg ), where ( m ) is the mass of the object and ( g ) is the acceleration due to gravity (( 9.8 , \text{m/s}^2 )).

So, ( N = 3 , \text{kg} \times 9.8 , \text{m/s}^2 = 29.4 , \text{N} ).

Now, we can find the minimum value of the coefficient of static friction by setting ( F_{\text{max}} = F_s ) and solving for ( \mu_s ).

( \mu_s \cdot N = 0.8 , \text{N} )

( \mu_s = \frac{0.8 , \text{N}}{29.4 , \text{N}} )

( \mu_s \approx 0.027 )

Therefore, the minimum value of the surface's coefficient of static friction is approximately ( 0.027 ).