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)
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)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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^](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb787237-32bd-4647-baa5-49b05320a55c%2F593b09bb-fc1a-4262-ba52-d38133c06216%2F76vf7tc_processed.png&w=3840&q=75)
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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