in the problem description it is suppose to be "Fdog up a hill" **Physics Problem: Box-Dog on an Inclined Plane**You and your dog, Box-Dog, go for a walk. In the scenario depicted in the image, Box-Dog is moving with a force of \( F_{dog} \) up a hill that has a coefficient of friction \( \mu \). The hill is inclined at an angle \( \theta \) and the leash makes an angle of \( \phi \) with respect to the normal, as shown.As the name suggests, you may treat Box-Dog as if he were a box. Suppose that Box-Dog moves up the hill with an acceleration of \( a = 0 \, \text{m/s}^2 \).**Diagram Description:**The diagram illustrates a box labeled "Box-Dog" on an inclined plane. The incline angle is marked as \( \theta \). The dog is represented as a box with a face, and an arrow labeled \( F_{dog} \) points up the slope, indicating the direction of force exerted on Box-Dog. The leash is shown extending from the box at an angle \( \phi \) to the normal of the hill.**Tasks:**d) **Using variables, solve for the tension in the leash.**e) **Using variables, solve for the normal force acting on Box-Dog.**To solve these tasks, consider forces acting parallel and perpendicular to the inclined plane. Use Newton's second law and the given conditions to find expressions for the tension and normal force, considering friction and the angles given.

d) Analyze the following figure. Let us take the direction of the box’s motion and vertically upward direction as the positive direction of motion. Applying Newton’s second law along the X-direction. Here, Fdog, T, m, f, ax, θ, and ϕ represent the box’s force, the leash’s tension, the box’s mass, friction, acceleration in X-direction, inclined plane’s angle, the leash’s angle with the Y-axis, respectively. As there is no acceleration along the X-direction, the equation of motion can be represented as, Here, N represents the normal force. Applying Newton’s second law along the Y-direction. Here, ay represents an acceleration along the Y-direction. As there is no motion along the Y-direction, the acceleration in this direction is zero. Thus,Form (1) and (2). The above expression on tension is the leash’s tension. e) Now, the normal force acting on the box can be found from equation (2) and (3) as, Thus, the above expression of N is the required normal force acting on the box.