19) A puck is kicked up a ramp, which makes an angle of $30^{\circ}$ with the horizontal. The graph below depicts the speed of the puck versus time. What is the coefficient of friction between the puck and the ramp?

A) 0.07
B) 0.15
C) 0.22
D) 0.29
E) 0.37
1. **Identify the given information and the goal:**
   - The ramp makes an angle of $30^\circ$ with the horizontal.
   - The graph depicts the speed of the puck versus time.
   - We need to find the coefficient of friction $\mu$ between the puck and the ramp.

2. **Analyze the graph:**
   - The x-intercept of the graph indicates the transition from the puck moving up the ramp to moving down.
   - The slope of the graph while the puck is moving up the ramp is $-2$ (indicating deceleration).
   - The slope of the graph while the puck is moving down the ramp is $-\frac{2}{3}$ (indicating acceleration).

3. **Determine the accelerations:**
   - The acceleration while the puck is moving up the ramp is $a_{\text{up}} = -2 \, \text{m/s}^2$.
   - The acceleration while the puck is moving down the ramp is $a_{\text{down}} = -\frac{2}{3} \, \text{m/s}^2$.

4. **Set up the equations for the accelerations:**
   - When the puck is moving up the ramp, the forces acting on it are gravity and friction. The net force is:
     \[
     F_{\text{net, up}} = -mg \sin(30^\circ) - \mu mg \cos(30^\circ)
     \]
     Therefore, the acceleration is:
     \[
     a_{\text{up}} = -g \sin(30^\circ) - \mu g \cos(30^\circ)
     \]
   - When the puck is moving down the ramp, the forces acting on it are gravity and friction. The net force is:
     \[
     F_{\text{net, down}} = mg \sin(30^\circ) - \mu mg \cos(30^\circ)
     \]
     Therefore, the acceleration is:
     \[
     a_{\text{down}} = g \sin(30^\circ) - \mu g \cos(30^\circ)
     \]

5. **Substitute the known values:**
   - $\sin(30^\circ) = \frac{1}{2}$
   - $\cos(30^\circ) = \frac{\sqrt{3}}{2}$
   - $g \approx 9.8 \, \text{m/s}^2$

   Thus, the equations become:
   \[
   a_{\text{up}} = -g \left( \frac{1}{2} \right) - \mu g \left( \frac{\sqrt{3}}{2} \right) = -4.9 - 4.9\sqrt{3}\mu
   \]
   \[
   a_{\text{down}} = g \left( \frac{1}{2} \right) - \mu g \left( \frac{\sqrt{3}}{2} \right) = 4.9 - 4.9\sqrt{3}\mu
   \]

6. **Set up the ratio of the accelerations:**
   \[
   \frac{a_{\text{up}}}{a_{\text{down}}} = \frac{-4.9 - 4.9\sqrt{3}\mu}{4.9 - 4.9\sqrt{3}\mu} = \frac{-2}{-\frac{2}{3}} = 3
   \]

7. **Solve for $\mu$:**
   \[
   \frac{-4.9 - 4.9\sqrt{3}\mu}{4.9 - 4.9\sqrt{3}\mu} = 3
   \]
   \[
   -4.9 - 4.9\sqrt{3}\mu = 3(4.9 - 4.9\sqrt{3}\mu)
   \]
   \[
   -4.9 - 4.9\sqrt{3}\mu = 14.7 - 14.7\sqrt{3}\mu
   \]
   \[
   -4.9 - 4.9\sqrt{3}\mu + 14.7\sqrt{3}\mu = 14.7
   \]
   \[
   -4.9 + 9.8\sqrt{3}\mu = 14.7
   \]
   \[
   9.8\sqrt{3}\mu = 19.6
   \]
   \[
   \mu = \frac{19.6}{9.8\sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \approx 0.29
   \]

The final answer is $\boxed{0.29}$