Suppose that 5 J of work is needed to stretch a spring from its natural length of 26 cm to a length of 40 cm.
(a) How much work is needed to stretch the spring from 28 cm to 32 cm? (Round your answer to two decimal places.)
0.82
 
Correct: Your answer is correct.
 J

(b) How far beyond its natural length will a force of 30 N keep the spring stretched? (Round your answer one decimal place.)
 cm

This problem can be solved using **Hooke's Law** and the work-energy principle. 

### Hooke's Law
The force required to stretch or compress a spring is proportional to the displacement \( x \) from its natural length:
\[
F(x) = kx
\]
where \( k \) is the spring constant and \( x \) is the displacement from the natural length.

### Work done to stretch a spring
The work \( W \) required to stretch a spring from displacement \( x_1 \) to \( x_2 \) is given by:
\[
W = \int_{x_1}^{x_2} F(x) \, dx = \int_{x_1}^{x_2} kx \, dx = \frac{k}{2} (x_2^2 - x_1^2)
\]
Where:
- \( W \) is the work,
- \( k \) is the spring constant,
- \( x_1 \) and \( x_2 \) are the displacements from the natural length.

### Part (a): Finding \( k \) (Spring constant)
From the information given, we know that 5 J of work is required to stretch the spring from 26 cm to 40 cm. This gives us the displacement \( x_1 = 0 \) and \( x_2 = 40 - 26 = 14 \) cm (or 0.14 m).

Using the work formula:
\[
W = \frac{k}{2} (x_2^2 - x_1^2)
\]
\[
5 = \frac{k}{2} (0.14^2)
\]
\[
5 = \frac{k}{2} \times 0.0196
\]
\[
k = \frac{5 \times 2}{0.0196} = 510.2 \, \text{N/m}
\]

### Part (b): Finding the displacement for 30 N of force
We now use Hooke's Law \( F(x) = kx \) to find the displacement \( x \) when the force is 30 N:
\[
F = kx
\]
\[
30 = 510.2 \times x
\]
\[
x = \frac{30}{510.2} = 0.0588 \, \text{m}
\]

This displacement is 0.0588 m or 5.88 cm.

Thus, the spring will be stretched **5.9 cm** beyond its natural length under a force of 30 N (rounded to one decimal place).

---

Would you like more details or have any questions about this solution? 

Here are some related questions for further exploration:
1. How does Hooke's Law apply to materials beyond springs?
2. What happens if the spring is stretched beyond its elastic limit?
3. How is the work formula derived from Hooke's Law?
4. Can Hooke's Law be applied to compressive forces in the same way?
5. How do spring constants vary between different materials and spring types?

**Tip:** The spring constant \( k \) is a key parameter, determining how stiff or soft the spring is. A larger \( k \) means a stiffer spring.

Suppose that 5 J of work is needed to stretch a spring from its natural length of 26 cm to a length of 40 cm.
(a) How much work is needed to stretch the spring from 28 cm to 32 cm? (Round your answer to two decimal places.)
(b) How far beyond its natural length will a force of 30 N keep the spring stretched? (Round your answer to one decimal place.)

Solve how much work is needed to stretch a spring from 28 cm to 32 cm, and determine the displacement under a 30 N force using Hooke's Law. This physics problem involves the work-energy principle and calculus-based integration. Suitable for high school students in Grades 11-12.

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