EXAMPLE 5.15 Work Required to Stretch a Spring GOAL Apply the graphical method of finding work. PROBLEM One end of a horizontal spring (k = 80.0 N/m) is held fixed while an external force is applied to the free end, stretching it slowly from A = 0 to xB = 4.00 cm. (a) Find the work done by the applied force on the spring. (b) Find the additional work done in stretching the spring from XB = 4.00 cm to xc = 7.00 cm. SOLUTION (A) Find the work from XA = Compute the area of the smaller triangle. cm to xB = 4.00 cm Fapp A (B) Find the work from xg = 4.00 cm to xc = 7.00 cm. Compute the area of the large triangle and subtract the area of the smaller triangle. Fapp (80.0 N/m) (x) B STRATEGY For part (a), simply find the area of the smaller triangle in the figure, using A = 1/2bh, one- half the base times the height. For part (b), the easiest way to find the additional work done from XB = 4.00 cm to xB = 7.00 cm is to find the area of the new, larger triangle and subtract the area of the smaller triangle. 4.00 7.00 (cm) A graph of the external force required to stretch a spring that obeys Hooke's law versus the elongation of the spring. W = 1/2kxg² = 1/2(80.0 N/m)(0.040 m)² = 0.064 J W = 1/2kxc² - W = 1/2kxB² W = 1/2 (80.0 N/m) (0.0700 m)2 - 0.0640 J = 0.196 J- 0.0640 J = 0.132 J
EXAMPLE 5.15 Work Required to Stretch a Spring GOAL Apply the graphical method of finding work. PROBLEM One end of a horizontal spring (k = 80.0 N/m) is held fixed while an external force is applied to the free end, stretching it slowly from A = 0 to xB = 4.00 cm. (a) Find the work done by the applied force on the spring. (b) Find the additional work done in stretching the spring from XB = 4.00 cm to xc = 7.00 cm. SOLUTION (A) Find the work from XA = Compute the area of the smaller triangle. cm to xB = 4.00 cm Fapp A (B) Find the work from xg = 4.00 cm to xc = 7.00 cm. Compute the area of the large triangle and subtract the area of the smaller triangle. Fapp (80.0 N/m) (x) B STRATEGY For part (a), simply find the area of the smaller triangle in the figure, using A = 1/2bh, one- half the base times the height. For part (b), the easiest way to find the additional work done from XB = 4.00 cm to xB = 7.00 cm is to find the area of the new, larger triangle and subtract the area of the smaller triangle. 4.00 7.00 (cm) A graph of the external force required to stretch a spring that obeys Hooke's law versus the elongation of the spring. W = 1/2kxg² = 1/2(80.0 N/m)(0.040 m)² = 0.064 J W = 1/2kxc² - W = 1/2kxB² W = 1/2 (80.0 N/m) (0.0700 m)2 - 0.0640 J = 0.196 J- 0.0640 J = 0.132 J
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![**Example 5.15: Work Required to Stretch a Spring**
**Goal:**
Apply the graphical method of finding work.
**Problem:**
One end of a horizontal spring (\(k = 80.0 \, \text{N/m}\)) is held fixed while an external force is applied to the free end, stretching it slowly from \(x_A = 0\) to \(x_B = 4.00 \, \text{cm}\).
(a) Find the work done by the applied force on the spring.
(b) Find the additional work done in stretching the spring from \(x_B = 4.00 \, \text{cm}\) to \(x_C = 7.00 \, \text{cm}\).
**Strategy:**
For part (a), simply find the area of the smaller triangle in the figure, using \(A = \frac{1}{2}bh\), one-half the base times the height. For part (b), the easiest way to find the additional work done from \(x_B = 4.00 \, \text{cm}\) to \(x_C = 7.00 \, \text{cm}\) is to find the area of the new, larger triangle and subtract the area of the smaller triangle.
**Solution:**
**(A)** Find the work from \(x_A = 0 \, \text{cm}\) to \(x_B = 4.00 \, \text{cm}\)
Compute the area of the smaller triangle.
\[
W = \frac{1}{2}kx_B^2 = \frac{1}{2}(80.0 \, \text{N/m})(0.040 \, \text{m})^2 = \boxed{0.064 \, \text{J}}
\]
**(B)** Find the work from \(x_B = 4.00 \, \text{cm}\) to \(x_C = 7.00 \, \text{cm}\).
Compute the area of the large triangle and subtract the area of the smaller triangle.
\[
W = \frac{1}{2}kx_C^2 - W = \frac{1}{2}kx_B^2
\]
\[
W = \frac{1}{2}(80.0 \, \text{N/m})(0.0700](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a2be487-a4f1-4704-8a71-b11163658c73%2Fbc3e9e41-e6e0-4a70-8804-fab00107b4cc%2Fsvx8d7_processed.png&w=3840&q=75)
Transcribed Image Text:**Example 5.15: Work Required to Stretch a Spring**
**Goal:**
Apply the graphical method of finding work.
**Problem:**
One end of a horizontal spring (\(k = 80.0 \, \text{N/m}\)) is held fixed while an external force is applied to the free end, stretching it slowly from \(x_A = 0\) to \(x_B = 4.00 \, \text{cm}\).
(a) Find the work done by the applied force on the spring.
(b) Find the additional work done in stretching the spring from \(x_B = 4.00 \, \text{cm}\) to \(x_C = 7.00 \, \text{cm}\).
**Strategy:**
For part (a), simply find the area of the smaller triangle in the figure, using \(A = \frac{1}{2}bh\), one-half the base times the height. For part (b), the easiest way to find the additional work done from \(x_B = 4.00 \, \text{cm}\) to \(x_C = 7.00 \, \text{cm}\) is to find the area of the new, larger triangle and subtract the area of the smaller triangle.
**Solution:**
**(A)** Find the work from \(x_A = 0 \, \text{cm}\) to \(x_B = 4.00 \, \text{cm}\)
Compute the area of the smaller triangle.
\[
W = \frac{1}{2}kx_B^2 = \frac{1}{2}(80.0 \, \text{N/m})(0.040 \, \text{m})^2 = \boxed{0.064 \, \text{J}}
\]
**(B)** Find the work from \(x_B = 4.00 \, \text{cm}\) to \(x_C = 7.00 \, \text{cm}\).
Compute the area of the large triangle and subtract the area of the smaller triangle.
\[
W = \frac{1}{2}kx_C^2 - W = \frac{1}{2}kx_B^2
\]
\[
W = \frac{1}{2}(80.0 \, \text{N/m})(0.0700
![**Educational Text Transcription and Explanation:**
**REMARKS**
Only simple geometries—rectangles and triangles—can be solved exactly with this method. More complex shapes require calculus or the square-counting technique in the next worked example.
**QUESTION**
When stretching springs, does half the displacement require half as much work? (Select all that apply.)
- [ ] Yes, because the work is equal to the force times the distance.
- [ ] Yes, because the spring exerts a constant force.
- [x] Yes, because half the displacement means the force from the spring is half as large.
- [ ] No, because the spring does not exert a constant force.
- [ ] No, because the force of the spring is proportional to \( x \).
- [x] No, because the force exerted by the spring is proportional to \( x^2 \).
---
**Considerations for Understanding the Work Done by an Applied Force**
*Consider the expression for the work done by an applied force. Under what conditions is the work done equal to simply a value of the force multiplied by the displacement? Do those conditions apply to the present problem - consider why or why not. Consider how the force exerted by the spring depends on its displacement, and what that implies about the expression for the work.*
---
**Explanation:**
This content encourages students to explore the relationship between displacement and work within the context of spring mechanics. It challenges students to consider how the work done is not only dependent on force but also on the proportionality to displacement, as indicated by Hooke's Law for springs. The checkboxes indicate different conceptual understandings of how spring force is proportional to either linear or squared displacement values. The questions and prompts guide learners to think critically about when simplifications like "work equals force times distance" hold true and under what physical conditions these might break down.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a2be487-a4f1-4704-8a71-b11163658c73%2Fbc3e9e41-e6e0-4a70-8804-fab00107b4cc%2F6ezf6i_processed.png&w=3840&q=75)
Transcribed Image Text:**Educational Text Transcription and Explanation:**
**REMARKS**
Only simple geometries—rectangles and triangles—can be solved exactly with this method. More complex shapes require calculus or the square-counting technique in the next worked example.
**QUESTION**
When stretching springs, does half the displacement require half as much work? (Select all that apply.)
- [ ] Yes, because the work is equal to the force times the distance.
- [ ] Yes, because the spring exerts a constant force.
- [x] Yes, because half the displacement means the force from the spring is half as large.
- [ ] No, because the spring does not exert a constant force.
- [ ] No, because the force of the spring is proportional to \( x \).
- [x] No, because the force exerted by the spring is proportional to \( x^2 \).
---
**Considerations for Understanding the Work Done by an Applied Force**
*Consider the expression for the work done by an applied force. Under what conditions is the work done equal to simply a value of the force multiplied by the displacement? Do those conditions apply to the present problem - consider why or why not. Consider how the force exerted by the spring depends on its displacement, and what that implies about the expression for the work.*
---
**Explanation:**
This content encourages students to explore the relationship between displacement and work within the context of spring mechanics. It challenges students to consider how the work done is not only dependent on force but also on the proportionality to displacement, as indicated by Hooke's Law for springs. The checkboxes indicate different conceptual understandings of how spring force is proportional to either linear or squared displacement values. The questions and prompts guide learners to think critically about when simplifications like "work equals force times distance" hold true and under what physical conditions these might break down.
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