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
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
Related questions
Topic Video
Question

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^
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 1 images

Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Recommended textbooks for you

College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning

University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON

Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press

College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning

University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON

Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press

Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning

Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley

College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON