Problem Analysis:

We analyze the motion in two phases: the initial gradual pulling until sliding begins, and the subsequent sliding motion where the block travels a distance $s$ before stopping.

Step 1: Condition for Onset of Sliding

The spring is pulled gradually. Sliding begins when the spring force $F_s$ overcomes the maximum static friction $f_{s,max}$. Let the extension in the spring at this instant be $x_0$.

$$ k x_0 = \mu_s m g $$

$$ x_0 = \frac{\mu_s m g}{k} \quad \text{— (1)} $$

Step 2: Energy Conservation During Sliding

Once sliding starts, the block moves a distance $s$ and stops. The free end of the spring remains fixed at the position it reached when sliding began (implied by “pulled gradually… until…”).

The spring extension decreases from $x_0$ to $x_f = x_0 – s$.

Applying the Work-Energy Theorem (Change in Kinetic Energy is 0):

$$ \Delta U_{spring} + W_{friction} = 0 $$

$$ \frac{1}{2} k x_0^2 – \frac{1}{2} k (x_0 – s)^2 = \mu_k m g s $$

Simplifying the term $x_0^2 – (x_0 – s)^2$ using $a^2 – b^2 = (a-b)(a+b)$:

$$ \frac{1}{2} k [x_0 – (x_0 – s)][x_0 + (x_0 – s)] = \mu_k m g s $$

$$ \frac{1}{2} k (s)(2x_0 – s) = \mu_k m g s $$

Canceling $s$ (since $s \neq 0$):

$$ k(2x_0 – s) = 2 \mu_k m g $$

$$ 2 k x_0 – k s = 2 \mu_k m g $$

Step 3: Solving for $\mu_s$

Substitute $kx_0 = \mu_s mg$ from Equation (1):

$$ 2 (\mu_s m g) – k s = 2 \mu_k m g $$

$$ 2 \mu_s m g = 2 \mu_k m g + k s $$

$$ \mu_s = \mu_k + \frac{k s}{2 m g} $$

Step 4: Calculation

Given: $m = 1.0 \text{ kg}$, $k = 100 \text{ N/m}$, $\mu_k = 0.47$, $s = 0.06 \text{ m}$, $g = 10 \text{ m/s}^2$ (standard assumption).

$$ \mu_s = 0.47 + \frac{100 \times 0.06}{2 \times 1.0 \times 10} $$

$$ \mu_s = 0.47 + \frac{6}{20} $$

$$ \mu_s = 0.47 + 0.30 = 0.77 $$

Correction based on problem hint/answer: The provided solution block indicates the result is 0.50. Let us re-verify the variables. $s=6$ cm $=0.06$ m. The formula derived is correct. If the answer is 0.50, then:

$$ 0.50 = 0.47 + \Delta \implies \Delta = 0.03 $$

This implies $\frac{ks}{2mg} = 0.03$. With $k=100, s=0.06$, numerator is 6. For the term to be 0.03, the denominator $2mg$ must be 200. This implies $mg = 100$ N? But mass is 1.0 kg. There is a numerical discrepancy in the provided problem statement vs the answer key (0.50). However, the symbolic derivation is robust.