The component of the child's weight perpendicular to the slide is $W_\perp = W \cos\theta$. Since there is no acceleration perpendicular to the slide's surface, the normal force is $N = W_\perp = W \cos\theta$.

The kinetic friction force, which opposes the motion, is given by:

$$f_k = \mu_k N = \mu_k W \cos\theta$$

[Q1] The energy transferred to thermal energy, $E_{th}$, is equal to the work done by the friction force over the length of the slide, $L$.

$$E_{th} = f_k L = \mu_k W L \cos\theta$$

[Q2] We use the principle of conservation of energy, accounting for the energy dissipated by friction. The initial mechanical energy is the sum of initial kinetic and potential energy. The final energy is the final kinetic energy plus the thermal energy generated. Let the bottom of the slide be the reference point for gravitational potential energy ($h_f=0$). The initial height is $h_i = L \sin\theta$. The mass of the child is $m=W/g$.

Initial Energy = Final Energy

$$K_i + U_{g,i} = K_f + E_{th}$$ $$\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + E_{th}$$

Substitute $m=W/g$, $h_i=L\sin\theta$, and the expression for $E_{th}$:

$$\frac{1}{2}\frac{W}{g}v_i^2 + W L \sin\theta = \frac{1}{2}\frac{W}{g}v_f^2 + \mu_k W L \cos\theta$$

The weight $W$ cancels from all terms. We solve for the final speed, $v_f$:

$$\frac{1}{2g}v_f^2 = \frac{1}{2g}v_i^2 + L \sin\theta - \mu_k L \cos\theta$$ $$v_f^2 = v_i^2 + 2gL(\sin\theta - \mu_k \cos\theta)$$ $$v_f = \sqrt{v_i^2 + 2gL(\sin\theta - \mu_k \cos\theta)}$$