The work-energy theorem states that the change in mechanical energy is equal to the work done by non-conservative forces, $W_{nc}$. The only non-conservative force is friction. The change in kinetic energy is $\Delta K = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2$. The change in gravitational potential energy is $\Delta U_g = 0 - mgh = -mgh$. The work done by the average frictional force $F_{friction}$ over a distance $d$ is $W_{nc} = -F_{friction}d$. Therefore, $\Delta K + \Delta U_g = W_{nc}$, which gives $\frac{1}{2}mv^2 - mgh = -F_{friction}d$. The energy dissipated by friction is $E_{diss} = mgh - \frac{1}{2}mv^2$. The average frictional force is $F_{friction} = E_{diss}/d$.

Case 1: Ground tangent at the bottom [Q1] The slide ends horizontally. Let $\theta$ be the angle subtended by the arc. From the geometry, a right triangle can be formed with the center of the circle, the top of the slide, and a point on the vertical radius. The sides are $R$, $R-h$, and a third side. Thus, $\cos\theta = \frac{R-h}{R}$. The length of the slide is the arc length $s = R\theta$.

$$s = R \arccos\left(\frac{R-h}{R}\right) = R \arccos\left(1 - \frac{h}{R}\right)$$

[Q2] The average frictional force $F_f$ is the energy dissipated divided by the slide length $s$.

$$F_f = \frac{mgh - \frac{1}{2}mv^2}{s} = \frac{mgh - \frac{1}{2}mv^2}{R \arccos\left(1 - \frac{h}{R}\right)}$$

Case 2: Vertical tangent at the top [Q3] The slide starts horizontally. Let $\alpha$ be the angle subtended by the arc. The vertical drop is $h$. From geometry, $\sin\alpha = \frac{h}{R}$. The new length of the slide is the arc length $s' = R\alpha$.

$$s' = R \arcsin\left(\frac{h}{R}\right)$$

[Q4] The average frictional force $F_f'$ is the energy dissipated divided by the new slide length $s'$. The energy dissipated is the same, as it depends only on $m, g, h,$ and $v$.

$$F_f' = \frac{mgh - \frac{1}{2}mv^2}{s'} = \frac{mgh - \frac{1}{2}mv^2}{R \arcsin\left(\frac{h}{R}\right)}$$