College Physics
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ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![**Transcription for Educational Website:**
### Kinematics and Dynamics of Curvilinear Motion
**a. Discuss, with diagrams, the kinematics and dynamics of an object undergoing curvilinear motion.**
*Explanation:*
Curvilinear motion refers to the movement of an object along a curved path. The motion can be described using kinematic equations that take into account the position, velocity, and acceleration vectors of the object. Dynamics involves the forces causing this motion, usually described through Newton's laws of motion. Diagrams can illustrate the curvature of the path, velocity vectors at different points, and forces acting on the object.
**b. A 5.0 kg object has a velocity (5.0i + 2.0j) m/s. If its velocity changes to (6.0i - 4.0j) m/s, what is the net external work done on the object?**
*Calculation:*
1. Calculate the initial kinetic energy (KE_initial):
\[ KE_{\text{initial}} = \frac{1}{2} m v_{\text{initial}}^2 \]
\[ v_{\text{initial}} = \sqrt{(5.0)^2 + (2.0)^2} = \sqrt{25 + 4} = \sqrt{29} \]
\[ KE_{\text{initial}} = \frac{1}{2} \times 5.0 \times 29 = 72.5 \, \text{J} \]
2. Calculate the final kinetic energy (KE_final):
\[ KE_{\text{final}} = \frac{1}{2} m v_{\text{final}}^2 \]
\[ v_{\text{final}} = \sqrt{(6.0)^2 + (-4.0)^2} = \sqrt{36 + 16} = \sqrt{52} \]
\[ KE_{\text{final}} = \frac{1}{2} \times 5.0 \times 52 = 130 \, \text{J} \]
3. Net external work done on the object is the change in kinetic energy:
\[ W_{\text{net}} = KE_{\text{final}} - KE_{\text{initial}} \]
\[ W_{\text{net}} = 130 \, \text{J} - 72.5 \, \text{J} = 57.5 \](https://content.bartleby.com/qna-images/question/2432691a-c29a-4a02-b85a-93b716867369/29fbe13b-5256-4baf-9919-29297ae1ae73/pqzliu9_processed.png)
Transcribed Image Text:**Transcription for Educational Website:**
### Kinematics and Dynamics of Curvilinear Motion
**a. Discuss, with diagrams, the kinematics and dynamics of an object undergoing curvilinear motion.**
*Explanation:*
Curvilinear motion refers to the movement of an object along a curved path. The motion can be described using kinematic equations that take into account the position, velocity, and acceleration vectors of the object. Dynamics involves the forces causing this motion, usually described through Newton's laws of motion. Diagrams can illustrate the curvature of the path, velocity vectors at different points, and forces acting on the object.
**b. A 5.0 kg object has a velocity (5.0i + 2.0j) m/s. If its velocity changes to (6.0i - 4.0j) m/s, what is the net external work done on the object?**
*Calculation:*
1. Calculate the initial kinetic energy (KE_initial):
\[ KE_{\text{initial}} = \frac{1}{2} m v_{\text{initial}}^2 \]
\[ v_{\text{initial}} = \sqrt{(5.0)^2 + (2.0)^2} = \sqrt{25 + 4} = \sqrt{29} \]
\[ KE_{\text{initial}} = \frac{1}{2} \times 5.0 \times 29 = 72.5 \, \text{J} \]
2. Calculate the final kinetic energy (KE_final):
\[ KE_{\text{final}} = \frac{1}{2} m v_{\text{final}}^2 \]
\[ v_{\text{final}} = \sqrt{(6.0)^2 + (-4.0)^2} = \sqrt{36 + 16} = \sqrt{52} \]
\[ KE_{\text{final}} = \frac{1}{2} \times 5.0 \times 52 = 130 \, \text{J} \]
3. Net external work done on the object is the change in kinetic energy:
\[ W_{\text{net}} = KE_{\text{final}} - KE_{\text{initial}} \]
\[ W_{\text{net}} = 130 \, \text{J} - 72.5 \, \text{J} = 57.5 \
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