1. Define the System and Forces
* System: The box
* Forces:
* Applied Force (F): 120 N, horizontal
* Gravity (mg): Acts vertically downward
* Normal Force (N): Acts perpendicular to the incline, balancing the component of gravity perpendicular to the incline.
* Component of Gravity Parallel to the Incline (mg sin θ): This component acts to oppose the applied force.
2. Free Body Diagram
Draw a free body diagram to visualize the forces acting on the box.
3. Resolve Forces
* Resolve gravity:
* The component of gravity parallel to the incline is mg sin θ.
* The component of gravity perpendicular to the incline is mg cos θ.
* Resolve the applied force:
* The component of the applied force parallel to the incline is F cos θ.
* The component of the applied force perpendicular to the incline is F sin θ.
4. Apply Newton's Second Law
* Newton's Second Law (along the incline): ΣF = ma
* Net force along the incline: F cos θ - mg sin θ = ma
5. Solve for Acceleration
* Substitute the given values: 120 N * cos(34°) - (7 kg * 9.8 m/s² * sin(34°)) = (7 kg) * a
* Calculate the acceleration (a).
6. Use Kinematics to Find Final Velocity
* Kinematics equation: v² = u² + 2as
* Initial velocity (u): 0 m/s (starts from rest)
* Distance (s): 15 m
* Acceleration (a): You calculated this in step 5.
* Solve for the final velocity (v).
Let's calculate the answers:
* Acceleration:
* 120 N * cos(34°) - (7 kg * 9.8 m/s² * sin(34°)) = (7 kg) * a
* a ≈ 2.95 m/s²
* Final Velocity:
* v² = 0² + 2 * 2.95 m/s² * 15 m
* v ≈ 9.49 m/s
Therefore, the final velocity of the box after being pushed 15 meters up the incline is approximately 9.49 m/s.
