When an elastic object , such as a coil spring or rubber band, is subjected to a force f, an increase in length , called a strain , occurs . Hooke's law states that force f is directly porportional to strain s. Suppose that a coil spring has a natural length of 4 feet and that a force of 60 pounds stretches the length to 6 feet. What magnitude of force , in pounds , would stretch the spring to a length of 7 feet ?

Options

A. 40
B. `51 3/7`
C. 70
D. 90

Answer

To solve the problem step by step, we will use Hooke's law, which states that the force \( F \) is directly proportional to the strain \( s \). The strain can be defined as the change in length of the spring from its natural length.

### Step-by-Step Solution:

1. **Identify the Natural Length and Initial Conditions:**
- The natural length of the spring is given as \( 4 \) feet.
- When a force of \( 60 \) pounds is applied, the length of the spring stretches to \( 6 \) feet.

2. **Calculate the Strain for the First Condition:**
- The change in length (strain) when the force is \( 60 \) pounds is:
\[ s_1 = \text{Final Length} - \text{Natural Length} = 6 \text{ feet} - 4 \text{ feet} = 2 \text{ feet} \]

3. **Set Up the Proportional Relationship:**
- According to Hooke's law, we can express the relationship between the forces and strains as:
\[ \frac{F_1}{F_2} = \frac{s_1}{s_2} \]
- Here, \( F_1 = 60 \) pounds, \( s_1 = 2 \) feet, and \( s_2 \) is the strain when the length is \( 7 \) feet.

4. **Calculate the Strain for the Second Condition:**
- The change in length when the length is \( 7 \) feet is:
\[ s_2 = 7 \text{ feet} - 4 \text{ feet} = 3 \text{ feet} \]

5. **Substitute the Known Values into the Proportional Relationship:**
- We have:
\[ \frac{60}{F_2} = \frac{2}{3} \]

6. **Cross-Multiply to Solve for \( F_2 \):**
- Cross-multiplying gives:
\[ 60 \cdot 3 = 2 \cdot F_2 \]
- This simplifies to:
\[ 180 = 2 \cdot F_2 \]

7. **Divide by 2 to Find \( F_2 \):**
- Dividing both sides by \( 2 \):
\[ F_2 = \frac{180}{2} = 90 \text{ pounds} \]

### Final Answer:
The magnitude of force required to stretch the spring to a length of \( 7 \) feet is \( 90 \) pounds.

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