Let's break down the problem based on the given image.

Problem Summary:

• A box of mass $m = 12.0 \, \text{kg}$ is being pushed at constant speed up a frictionless ramp by a horizontal force $F$.
• The ramp is inclined at an angle $\theta = 36.0^\circ$ with respect to the horizontal.

Part A: Free-Body Diagram

The forces acting on the box are:

1. Gravitational force $\mathbf{W} = m \mathbf{g}$, acting vertically downward.
2. Normal force $\mathbf{N}$, acting perpendicular to the surface of the ramp.
3. Applied force $\mathbf{F}$, acting horizontally.

You would typically draw these forces with the following components:

• Weight splits into two components: one along the ramp $W_\parallel = mg \sin(\theta)$ and one perpendicular to the ramp $W_\perp = mg \cos(\theta)$.
• Normal force $N$ balances the perpendicular component of the weight.
• Horizontal force $F$ has components parallel and perpendicular to the incline, since it is applied horizontally but the motion is along the incline.

Part B: Equations of Motion for $m$

Since the box is moving at constant speed, the net force along the incline is zero, which means:

$F_\parallel = W_\parallel$

Let’s write out the components:

1. Along the incline: $F \cos(\theta) = mg \sin(\theta)$ This equation balances the horizontal force's component along the incline with the component of gravity pulling the box downward along the incline.

2. Perpendicular to the incline: $N + F \sin(\theta) = mg \cos(\theta)$ This equation ensures the normal force balances the perpendicular components of gravity and the applied force.

Part C: Magnitude of the Force $F$

From the first equation in Part B: $F \cos(\theta) = mg \sin(\theta)$

Solve for $F$: $F = \frac{mg \sin(\theta)}{\cos(\theta)}$

Plugging in the values: $F = \frac{(12.0 \, \text{kg})(9.8 \, \text{m/s}^2) \sin(36.0^\circ)}{\cos(36.0^\circ)}$

We can calculate the result shortly.

Part D: Magnitude of the Normal Force

From the second equation in Part B: $N = mg \cos(\theta) - F \sin(\theta)$

Substitute the expression for $F$ from Part C into this equation to get the normal force $N$.

Let's compute both the force $F$ and normal force $N$.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, I can guide you on how to calculate these values manually if you'd like to proceed:

1. Use the formula for $F$: $F = \frac{(12.0 \, \text{kg})(9.8 \, \text{m/s}^2) \sin(36.0^\circ)}{\cos(36.0^\circ)}$ First, calculate $\sin(36.0^\circ)$ and $\cos(36.0^\circ)$ using a calculator.

2. After finding $F$, substitute it into the equation for the normal force $N$: $N = mg \cos(\theta) - F \sin(\theta)$

Let me know if you want me to walk through the steps in more detail, or if you have any questions!

Here are five related questions:

1. How would the result change if the ramp had friction?
2. What happens if the box is not moving at constant speed?
3. Can you generalize the equation of motion for any incline angle?
4. How would the free-body diagram differ with an upward or downward applied force?
5. What if the box is pushed down the ramp instead of up?

Tip: When dealing with inclined planes, always break forces into parallel and perpendicular components to simplify the analysis.