Set the potential energy reference $U=0$ at the lowest point B.

[Q1] Apply the conservation of mechanical energy between A and B, where the path is frictionless.

$$K_A + U_A = K_B + U_B$$ $$\frac{1}{2}mv_A^2 + mgh_1 = \frac{1}{2}mv_B^2 + 0$$

Solving for $v_B$:

$$v_B = \sqrt{v_A^2 + 2gh_1}$$ $v_B = \sqrt{(7.0 \text{ m/s})^2 + 2(9.8 \text{ m/s}^2)(6.0 \text{ m})} = 12.9$ m/s.

[Q2] Apply the conservation of mechanical energy between A and C, which is also a frictionless path.

$$K_A + U_A = K_C + U_C$$ $$\frac{1}{2}mv_A^2 + mgh_1 = \frac{1}{2}mv_C^2 + mgh_2$$

Solving for $v_C$:

$$v_C = \sqrt{v_A^2 + 2g(h_1 - h_2)}$$ $v_C = \sqrt{(7.0 \text{ m/s})^2 + 2(9.8 \text{ m/s}^2)(6.0 \text{ m} - 2.0 \text{ m})} = 11.3$ m/s.

[Q3] On the rough section from C, friction does negative work $W_f = -f_k d = -\mu_k mgd$. To find the stopping distance $d$, we use the work-energy theorem. The block's kinetic energy at C is converted into thermal energy by friction.

$$\Delta E_{mech} = W_f$$ $$(K_{final} - K_C) + (U_{final} - U_C) = -\mu_k mgd$$

At stopping, $K_{final}=0$. Since the path is horizontal, $U_{final} = U_C$.

$$0 - \frac{1}{2}mv_C^2 = -\mu_k mgd$$ $$d = \frac{v_C^2}{2\mu_k g}$$

Using $v_C^2 = 127.4 \text{ m}^2/\text{s}^2$ from part (b):

$d = \frac{127.4 \text{ m}^2/\text{s}^2}{2(0.70)(9.8 \text{ m/s}^2)} = 9.3$ m.

Since $d = 9.3$ m is less than $L = 12$ m, the block does not reach point D.