A 1.5-kg object moving along the x axis has a velocity of +4.0 m/s at x = 0. If the only force acting on this object is shown in the figure, what is the kinetic energy of the object at x = +3.0 m? 8 4 0 Fx (N) 1 3 x (m)

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### Problem:
A 1.5-kg object moving along the x-axis has a velocity of +4.0 m/s at x = 0. If the only force acting on this object is shown in the figure, what is the kinetic energy of the object at x = +3.0 m?

### Given:
- Mass of the object (m): 1.5 kg
- Initial velocity (v₀): +4.0 m/s at x = 0

### Provided Figure Description:
The figure shows a force (Fₓ) vs. position (x) graph:
- The y-axis represents the force (Fₓ) in Newtons (N) ranging from -4 N to +8 N.
- The x-axis represents the position (x) in meters (m) ranging from -1 m to +4 m.

The graph consists of a straight line starting at:
- Fₓ = 8 N at x = 0 m 
- Decreases linearly to 
- Fₓ = -4 N at x = 4 m

### Steps to Solve:

1. **Determine the work done on the object:**
   The work done (W) by the force as the object moves from x = 0 to x = 3 m can be found by calculating the area under the Fₓ vs. x graph.

   - Break the area into geometric shapes (triangle and rectangle).
   
   Since the graph is a straight line, we can find the area under the graph between x = 0 and x = 3 m. 
   
   **For the triangular part (x = 0 to x = 3 m):**
   - Base = 3 m
   - Height = 8 N

   Area = 0.5 * base * height = 0.5 * 3 m * 8 N = 12 J

2. **Relate Work to Change in Kinetic Energy:**
   According to the work-energy theorem, the net work done on the object is equal to its change in kinetic energy (ΔK):

   \( W = \Delta K = K_{\text{final}} - K_{\text{initial}} \)
   
   Where \( K = \frac{1}{2}mv^2 \).

   - Initial Kinetic Energy, \( K_{\text{initial}} = \frac{1}{2} m v_0^
Transcribed Image Text:### Problem: A 1.5-kg object moving along the x-axis has a velocity of +4.0 m/s at x = 0. If the only force acting on this object is shown in the figure, what is the kinetic energy of the object at x = +3.0 m? ### Given: - Mass of the object (m): 1.5 kg - Initial velocity (v₀): +4.0 m/s at x = 0 ### Provided Figure Description: The figure shows a force (Fₓ) vs. position (x) graph: - The y-axis represents the force (Fₓ) in Newtons (N) ranging from -4 N to +8 N. - The x-axis represents the position (x) in meters (m) ranging from -1 m to +4 m. The graph consists of a straight line starting at: - Fₓ = 8 N at x = 0 m - Decreases linearly to - Fₓ = -4 N at x = 4 m ### Steps to Solve: 1. **Determine the work done on the object:** The work done (W) by the force as the object moves from x = 0 to x = 3 m can be found by calculating the area under the Fₓ vs. x graph. - Break the area into geometric shapes (triangle and rectangle). Since the graph is a straight line, we can find the area under the graph between x = 0 and x = 3 m. **For the triangular part (x = 0 to x = 3 m):** - Base = 3 m - Height = 8 N Area = 0.5 * base * height = 0.5 * 3 m * 8 N = 12 J 2. **Relate Work to Change in Kinetic Energy:** According to the work-energy theorem, the net work done on the object is equal to its change in kinetic energy (ΔK): \( W = \Delta K = K_{\text{final}} - K_{\text{initial}} \) Where \( K = \frac{1}{2}mv^2 \). - Initial Kinetic Energy, \( K_{\text{initial}} = \frac{1}{2} m v_0^
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