Here is the solution to the problem. The problem statement contains a contradiction, stating the ball rolls "without slipping" and then "with slipping". Based on the questions asked, the intended condition is rolling without slipping.

[Q1] Friction Requirement and Work Done

Friction Requirement: Rolling without slipping is impossible on a frictionless surface. A torque is required to initiate angular acceleration from rest. This torque is provided by static friction.

To find the minimum static friction coefficient $\mu_s$, we apply Newton's second law for translation and rotation.

1. Translational motion along the incline: $$mg\sin\theta - f_s = ma$$
2. Rotational motion about the center of mass (CM): $$\tau = f_s r = I\alpha$$
3. No-slip condition: $a = \alpha r$.

From (2) and (3), we find the required static friction force:

$f_s = I(a/r^2)$.

Substitute this into (1) to find the acceleration $a$:

$$mg\sin\theta - \frac{I}{r^2}a = ma \implies a = \frac{mg\sin\theta}{m+I/r^2} = \frac{g\sin\theta}{1+I/(mr^2)}$$

For a solid ball, the moment of inertia is $I = \frac{2}{5}mr^2$.

$$a = \frac{g\sin\theta}{1+2/5} = \frac{5}{7}g\sin\theta$$

The required friction force is:

$$f_s = \frac{I}{r^2}a = \left(\frac{2}{5}m\right)\left(\frac{5}{7}g\sin\theta\right) = \frac{2}{7}mg\sin\theta$$

For no slipping, this force must be less than or equal to the maximum static friction, $f_s \le \mu_s N$. The normal force is $N = mg\cos\theta$.

$$\frac{2}{7}mg\sin\theta \le \mu_s mg\cos\theta$$ $$\mu_s \ge \frac{2}{7}\frac{\sin\theta}{\cos\theta} \implies \mu_{s, \text{min}} = \frac{2}{7}\tan\theta$$

Work Done by Friction: The work done by static friction is zero. The point of application of the static friction force on the ball is instantaneously at rest relative to the slope. Since the displacement of the point of application is zero, the work done is zero ($W = \vec{F} \cdot d\vec{s} = 0$).

[Q2] Final Speed and Time

Method 1: Kinematics The acceleration is constant, $a = \frac{5}{7}g\sin\theta$. The distance traveled along the slope is $s = h/\sin\theta$. Using the kinematic equation $v^2 = v_0^2 + 2as$ with $v_0=0$:

$$v^2 = 2\left(\frac{5}{7}g\sin\theta\right)\left(\frac{h}{\sin\theta}\right) = \frac{10}{7}gh$$ $$v = \sqrt{\frac{10}{7}gh}$$

The time taken is found using $s = \frac{1}{2}at^2$:

$$t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2(h/\sin\theta)}{(5/7)g\sin\theta}} = \sqrt{\frac{14h}{5g\sin^2\theta}} = \frac{1}{\sin\theta}\sqrt{\frac{14h}{5g}}$$

Method 2: Energy Conservation Since static friction does no work, mechanical energy is conserved. Initial Energy ($E_i$) = Final Energy ($E_f$)

$$U_i + K_i = U_f + K_f$$ $$mgh + 0 = 0 + K_{trans} + K_{rot}$$ $$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Using the no-slip condition $\omega = v/r$ and $I = \frac{2}{5}mr^2$:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2$$ $$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$$

Solving for the final speed $v$:

$$v^2 = \frac{10}{7}gh \implies v = \sqrt{\frac{10}{7}gh}$$

To find the time, we must use kinematics with the acceleration derived previously, which yields the same result as Method 1.

[Q3] Comparison of Final Speeds

The final speed for any object rolling without slipping can be expressed generally using the energy conservation result:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}v^2(m + I/r^2)$$ $$v = \sqrt{\frac{2gh}{1 + I/(mr^2)}}$$

Let's compare this for three objects by defining $I = cmr^2$:

1. Solid Ball: $I = \frac{2}{5}mr^2 \implies c = 2/5$. $$v_{ball} = \sqrt{\frac{2gh}{1+2/5}} = \sqrt{\frac{10}{7}gh} \approx \sqrt{1.43gh}$$
2. Solid Disk: $I = \frac{1}{2}mr^2 \implies c = 1/2$. $$v_{disk} = \sqrt{\frac{2gh}{1+1/2}} = \sqrt{\frac{4}{3}gh} \approx \sqrt{1.33gh}$$
3. Thin Ring: $I = mr^2 \implies c = 1$. $$v_{ring} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$$

Comparing the final speeds:

$$v_{ball} > v_{disk} > v_{ring}$$

The ball is the fastest because it has the smallest moment of inertia, requiring less of its potential energy to be converted into rotational kinetic energy, leaving more for translational kinetic energy.

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