Page 138 Practice Problem 5.14: If, instead of the fish, we place a 3.00 kg rock in the pan, how much does the spring stretch? Answer: twice as much, 2.45 cm.
Page 138 Practice Problem 5.14: If, instead of the fish, we place a 3.00 kg rock in the pan, how much does the spring stretch? Answer: twice as much, 2.45 cm.
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)...
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The second image has background info used to help solve the practice problem.
![**Chapter 5: Applications of Newton’s Laws**
### Example 5.4: Fishy Business
In this example, we will consider the forces that come into play in measuring weight. Weighing a fish involves using a spring balance. Let's go through a scenario:
- **Set Up**:
When the scale is unstressed, both the spring and the fish provide a reading of zero. In Figure 5.22a and b, a 1.50-kg fish is weighed.
**Solution**:
1. **Weight Calculation**:
- The weight of the fish is:
\[
mg = (1.50\, \text{kg})(9.80\, \text{m/s}^2) = 14.7\, \text{N}
\]
2. **Equilibrium Condition**:
- \[
\sum F = F_{\text{sp}} - mg = 0 \\
\Rightarrow F_{\text{sp}} = mg = 14.7\, \text{N}
\]
3. **Spring Calculation**:
- The spring constant \( k \) is given as 1200 N/m.
- Use Hooke's Law:
\[
F = k \Delta L = 14.7\, \text{N} \\
\Delta L = \frac{F}{k} = \frac{14.7\, \text{N}}{1200\, \text{N/m}} = 0.0123\, \text{m} = 1.23\, \text{cm}
\]
- **Reflect**:
- The weight of the 1.50-kg fish is less than the tested weight of 12.0 N, so the spring stretches less than the tested 10 cm, specifically about 1.23 cm.
- **Practice Problem**:
- If a 3.00 kg rock is placed on the scale instead, calculate the spring stretch.
- **Answer**: The spring stretches twice as much, or about 2.45 cm.
**Figure 5.22: Weighing a Fish**
1. **Diagram (a):** Shows the spring stretched by a known weight, 12.0 N.
2. **Diagram (b):** Illustrates the spring stretched by the fish.
3. **Diagram (c](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F28764828-4d1c-44b8-9544-7d79c8b69e6b%2F1c9fdbff-ea60-4368-83a6-e0743fbf3893%2Fjpx7awe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chapter 5: Applications of Newton’s Laws**
### Example 5.4: Fishy Business
In this example, we will consider the forces that come into play in measuring weight. Weighing a fish involves using a spring balance. Let's go through a scenario:
- **Set Up**:
When the scale is unstressed, both the spring and the fish provide a reading of zero. In Figure 5.22a and b, a 1.50-kg fish is weighed.
**Solution**:
1. **Weight Calculation**:
- The weight of the fish is:
\[
mg = (1.50\, \text{kg})(9.80\, \text{m/s}^2) = 14.7\, \text{N}
\]
2. **Equilibrium Condition**:
- \[
\sum F = F_{\text{sp}} - mg = 0 \\
\Rightarrow F_{\text{sp}} = mg = 14.7\, \text{N}
\]
3. **Spring Calculation**:
- The spring constant \( k \) is given as 1200 N/m.
- Use Hooke's Law:
\[
F = k \Delta L = 14.7\, \text{N} \\
\Delta L = \frac{F}{k} = \frac{14.7\, \text{N}}{1200\, \text{N/m}} = 0.0123\, \text{m} = 1.23\, \text{cm}
\]
- **Reflect**:
- The weight of the 1.50-kg fish is less than the tested weight of 12.0 N, so the spring stretches less than the tested 10 cm, specifically about 1.23 cm.
- **Practice Problem**:
- If a 3.00 kg rock is placed on the scale instead, calculate the spring stretch.
- **Answer**: The spring stretches twice as much, or about 2.45 cm.
**Figure 5.22: Weighing a Fish**
1. **Diagram (a):** Shows the spring stretched by a known weight, 12.0 N.
2. **Diagram (b):** Illustrates the spring stretched by the fish.
3. **Diagram (c
Transcribed Image Text:**Practice Problem 5.14:**
If, instead of the fish, we place a 3.00 kg rock in the pan, how much does the spring stretch?
**Answer:** twice as much, 2.45 cm.
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