Answer :
Sure! Let's find the work done to stretch a spring from its relaxed state to a total of 3 inches, given the information provided:
### Step 1: Determine the spring constant [tex]\( k \)[/tex]
When a spring stretches according to Hooke's Law, the force [tex]\( F \)[/tex] is directly proportional to the distance [tex]\( x \)[/tex] it is stretched. The constant of proportionality is known as the spring constant [tex]\( k \)[/tex].
Given:
- [tex]\( F = 180 \)[/tex] pounds
- [tex]\( x = 2 \)[/tex] inches
Hooke's Law is expressed as:
[tex]\[ F = k \cdot x \][/tex]
We can solve for the spring constant [tex]\( k \)[/tex] by rearranging the equation:
[tex]\[ k = \frac{F}{x} \][/tex]
Substituting the given values:
[tex]\[ k = \frac{180 \text{ pounds}}{2 \text{ inches}} = 90 \text{ pounds per inch} \][/tex]
### Step 2: Calculate the work done to stretch the spring
The work [tex]\( W \)[/tex] done in stretching or compressing a spring from its relaxed state is given by:
[tex]\[ W = \frac{1}{2} k x^2 \][/tex]
We need to find the work done to stretch the spring to a total of 3 inches. Given that [tex]\( k = 90 \)[/tex] pounds per inch, we substitute [tex]\( k \)[/tex] and [tex]\( x = 3 \)[/tex] inches into the work formula:
[tex]\[ W = \frac{1}{2} \times 90 \text{ pounds/inch} \times (3 \text{ inches})^2 \][/tex]
First, calculate [tex]\( (3 \text{ inches})^2 \)[/tex]:
[tex]\[ (3 \text{ inches})^2 = 9 \text{ square inches} \][/tex]
Now substitute back into the work formula:
[tex]\[ W = \frac{1}{2} \times 90 \times 9 \][/tex]
Multiply the values:
[tex]\[ W = \frac{1}{2} \times 810 \][/tex]
[tex]\[ W = 405 \][/tex]
Therefore, the work done in stretching the spring from its relaxed state to a total of 3 inches is [tex]\( 405 \)[/tex] inch-pounds.
### Step 1: Determine the spring constant [tex]\( k \)[/tex]
When a spring stretches according to Hooke's Law, the force [tex]\( F \)[/tex] is directly proportional to the distance [tex]\( x \)[/tex] it is stretched. The constant of proportionality is known as the spring constant [tex]\( k \)[/tex].
Given:
- [tex]\( F = 180 \)[/tex] pounds
- [tex]\( x = 2 \)[/tex] inches
Hooke's Law is expressed as:
[tex]\[ F = k \cdot x \][/tex]
We can solve for the spring constant [tex]\( k \)[/tex] by rearranging the equation:
[tex]\[ k = \frac{F}{x} \][/tex]
Substituting the given values:
[tex]\[ k = \frac{180 \text{ pounds}}{2 \text{ inches}} = 90 \text{ pounds per inch} \][/tex]
### Step 2: Calculate the work done to stretch the spring
The work [tex]\( W \)[/tex] done in stretching or compressing a spring from its relaxed state is given by:
[tex]\[ W = \frac{1}{2} k x^2 \][/tex]
We need to find the work done to stretch the spring to a total of 3 inches. Given that [tex]\( k = 90 \)[/tex] pounds per inch, we substitute [tex]\( k \)[/tex] and [tex]\( x = 3 \)[/tex] inches into the work formula:
[tex]\[ W = \frac{1}{2} \times 90 \text{ pounds/inch} \times (3 \text{ inches})^2 \][/tex]
First, calculate [tex]\( (3 \text{ inches})^2 \)[/tex]:
[tex]\[ (3 \text{ inches})^2 = 9 \text{ square inches} \][/tex]
Now substitute back into the work formula:
[tex]\[ W = \frac{1}{2} \times 90 \times 9 \][/tex]
Multiply the values:
[tex]\[ W = \frac{1}{2} \times 810 \][/tex]
[tex]\[ W = 405 \][/tex]
Therefore, the work done in stretching the spring from its relaxed state to a total of 3 inches is [tex]\( 405 \)[/tex] inch-pounds.
