To solve this problem, we need to apply the principles of energy conservation and the equations of projectile motion. Here’s the breakdown of the solution:

Given Data

1. Mass of the box, $m = 0.450 \, \text{kg}$
2. Spring compression, $x = 0.198 \, \text{m}$
3. Spring constant, $k = 1350 \, \text{N/m}$
4. Distance along the incline, $d = 1.65 \, \text{m}$
5. Angle of incline, $\theta = 42.5^\circ$
6. Maximum height of the projectile, $H = 3.25 \, \text{m}$
7. Coefficient of friction on the incline, $\mu$ (to be determined)

Solution Steps

1. Energy Stored in the Spring: The initial energy in the spring is given by: $E_{\text{spring}} = \frac{1}{2}kx^2$ Substituting the values: $E_{\text{spring}} = \frac{1}{2} \times 1350 \times (0.198)^2$ Calculate $E_{\text{spring}}$.

2. Energy at the Top of the Incline (before the box leaves the ramp): By the work-energy theorem, the initial energy stored in the spring is used to overcome:

   • The gravitational potential energy gain along the incline
   • The work done against friction

   Let $E_{\text{top}}$ be the kinetic energy at the top of the incline. We have: $E_{\text{spring}} = m g d \sin(\theta) + \mu m g d \cos(\theta) + E_{\text{top}}$

   Rearranging to solve for $E_{\text{top}}$: $E_{\text{top}} = E_{\text{spring}} - m g d \sin(\theta) - \mu m g d \cos(\theta)$

3. Projectile Motion Analysis: At the top of the incline, the box transitions into projectile motion. The vertical height $H$ reached is given by the initial vertical velocity component squared divided by $2g$: $H = \frac{(v_y)^2}{2g}$ where $v_y = v \sin(\theta)$ and $v$ is the speed at the top of the incline.

4. Calculating $\mu$: By substituting the values for $H$, solving for $v_y$ and using energy equations, we can find