A block of mass $\text{m = 3.0 kg}$ slides down one ramp, and then up a second ramp. The coefficient of kinetic friction between the block and each ramp is $\mu_\text{k} = \text{0.40}$. The block begins at a height $\text{h}_\text{1} = \text{1.0 m}$ above the horizontal. Both ramps are at a $\text{30}^{\circ}$ incline above the horizontal. To what height above the horizontal does the block rise on the second ramp?

(A) $\text{0.18 m}$
(B) $\text{0.52 m}$
(C) $\text{0.59 m}$
(D) $\text{0.69 m}$
(E) $\text{0.71 m}$
1. **Understanding the problem:**
   - A block of mass \( m = 3.0 \, \text{kg} \) slides down a ramp and then up another ramp.
   - The coefficient of kinetic friction between the block and each ramp is \( \mu_k = 0.40 \).
   - The block starts at a height \( h_1 = 1.0 \, \text{m} \) above the horizontal.
   - Both ramps are inclined at \( 30^\circ \) above the horizontal.
   - We need to find the height \( h_2 \) above the horizontal that the block reaches on the second ramp.

2. **Applying the conservation of energy:**
   - The initial potential energy of the block at height \( h_1 \) is \( mgh_1 \).
   - As the block slides down the first ramp and up the second ramp, it loses energy due to friction.
   - The energy lost due to friction is given by \( \mu_k F_N L \), where \( F_N \) is the normal force and \( L \) is the distance traveled along the ramp.

3. **Calculating the normal force and distance:**
   - The normal force \( F_N \) on an inclined plane is \( F_N = mg \cos \theta \).
   - The distance \( L \) traveled along the ramp can be found using trigonometry: \( L = \frac{h_2}{\sin \theta} \).

4. **Setting up the energy conservation equation:**
   - The initial energy is \( mgh_1 \).
   - The final energy is \( mgh_2 \) plus the energy lost to friction.
   - The energy lost to friction is \( \mu_k mg \cos \theta \cdot \frac{h_2}{\sin \theta} \).

   Therefore, the energy conservation equation is:
   \[
   mgh_1 = mgh_2 + \mu_k mg \cos \theta \cdot \frac{h_2}{\sin \theta}
   \]

5. **Simplifying the equation:**
   - Cancel out the common factor \( mg \) from both sides:
   \[
   h_1 = h_2 + \mu_k \cos \theta \cdot \frac{h_2}{\sin \theta}
   \]

   - Substitute \( \cos \theta = \cos 30^\circ = \frac{\sqrt{3}}{2} \) and \( \sin \theta = \sin 30^\circ = \frac{1}{2} \):
   \[
   h_1 = h_2 + \mu_k \cdot \frac{\sqrt{3}}{2} \cdot \frac{h_2}{\frac{1}{2}}
   \]

   - Simplify the expression:
   \[
   h_1 = h_2 + \mu_k \cdot \sqrt{3} \cdot h_2
   \]

   - Factor out \( h_2 \):
   \[
   h_1 = h_2 (1 + \mu_k \sqrt{3})
   \]

6. **Solving for \( h_2 \):**
   - Substitute \( h_1 = 1.0 \, \text{m} \) and \( \mu_k = 0.40 \):
   \[
   1.0 = h_2 (1 + 0.40 \cdot \sqrt{3})
   \]

   - Calculate the numerical value:
   \[
   1.0 = h_2 (1 + 0.40 \cdot 1.732)
   \]
   \[
   1.0 = h_2 (1 + 0.6928)
   \]
   \[
   1.0 = h_2 \cdot 1.6928
   \]
   \[
   h_2 = \frac{1.0}{1.6928} \approx 0.59 \, \text{m}
   \]

The final answer is \( \boxed{0.59 \, \text{m}} \)