Answer :
To solve this problem, we need to first understand the physics of a spring and how work is related to its stretching.
A spring follows Hooke's Law, which states that the force exerted by a spring is proportional to its displacement from the equilibrium position, or:
[tex]F = kx[/tex]
where:
- [tex]F[/tex] is the force exerted by the spring in Newtons (N),
- [tex]k[/tex] is the spring constant, and
- [tex]x[/tex] is the displacement from the unstretched position in meters (m).
The work done on a spring when it's stretched or compressed is given by:
[tex]W = \frac{1}{2} k x^2[/tex]
where:
- [tex]W[/tex] is the work done (in Joules),
- [tex]k[/tex] is the spring constant (in N/m),
- [tex]x[/tex] is the displacement (in m).
Part 1: Finding Final Length of the Spring
Given:
- The spring constant [tex]k = 150 \, \text{N/m}[/tex].
- The work done [tex]W = 220 \, \text{J}[/tex].
Plug these values into the work equation to find the displacement [tex]x[/tex]:
[tex]220 = \frac{1}{2} \times 150 \times x^2[/tex]
Simplifying gives:
[tex]220 = 75x^2[/tex]
[tex]x^2 = \frac{220}{75}[/tex]
[tex]x^2 \approx 2.933[/tex]
[tex]x \approx \sqrt{2.933}[/tex]
[tex]x \approx 1.712 \, \text{m}[/tex]
The original length of the spring is 0.4 m. Therefore, the final length when stretched will be:
[tex]\text{Final length} = 0.4 \text{ m} + 1.712 \text{ m} = 2.112 \text{ m}[/tex]
Part 2: Finding the Magnitude of Force at Maximum Elongation
To find the force at maximum elongation, use Hooke's Law [tex]F = kx[/tex]:
[tex]F = 150 \times 1.712[/tex]
[tex]F \approx 256.8 \, \text{N}[/tex]
So, at maximum elongation, the force [tex]F[/tex] is approximately 256.8 N.
In summary, the final length of the spring when stretched is approximately 2.112 m, and the magnitude of the force at maximum elongation is approximately 256.8 N.