Answer :
To find out how much the spring should stretch to store 360 Joules of elastic potential energy, we can use the formula for elastic potential energy:
[tex]\[ PE = \frac{1}{2} K \cdot X^2 \][/tex]
where:
- [tex]\( PE \)[/tex] is the potential energy in Joules.
- [tex]\( K \)[/tex] is the spring constant in N/m.
- [tex]\( X \)[/tex] is the displacement or stretch in meters.
Given that the spring constant [tex]\( K = 423.25 \, \text{N/m} \)[/tex] and the potential energy [tex]\( PE = 360 \, \text{J} \)[/tex], we can substitute these values into the equation and solve for [tex]\( X \)[/tex].
1. First, multiply both sides of the equation by 2 to get rid of the fraction:
[tex]\[ 2 \times PE = K \cdot X^2 \][/tex]
2. Plug the given values into the equation:
[tex]\[ 2 \times 360 = 423.25 \cdot X^2 \][/tex]
3. Solve for [tex]\( X^2 \)[/tex]:
[tex]\[ X^2 = \frac{720}{423.25} \][/tex]
4. Calculate [tex]\( X^2 \)[/tex]:
[tex]\[ X^2 \approx 1.701 \][/tex]
5. Finally, take the square root of both sides to find [tex]\( X \)[/tex]:
[tex]\[ X \approx \sqrt{1.701} \approx 1.304 \][/tex]
Therefore, the spring must stretch approximately 1.304 meters to store 360 Joules of elastic potential energy.
[tex]\[ PE = \frac{1}{2} K \cdot X^2 \][/tex]
where:
- [tex]\( PE \)[/tex] is the potential energy in Joules.
- [tex]\( K \)[/tex] is the spring constant in N/m.
- [tex]\( X \)[/tex] is the displacement or stretch in meters.
Given that the spring constant [tex]\( K = 423.25 \, \text{N/m} \)[/tex] and the potential energy [tex]\( PE = 360 \, \text{J} \)[/tex], we can substitute these values into the equation and solve for [tex]\( X \)[/tex].
1. First, multiply both sides of the equation by 2 to get rid of the fraction:
[tex]\[ 2 \times PE = K \cdot X^2 \][/tex]
2. Plug the given values into the equation:
[tex]\[ 2 \times 360 = 423.25 \cdot X^2 \][/tex]
3. Solve for [tex]\( X^2 \)[/tex]:
[tex]\[ X^2 = \frac{720}{423.25} \][/tex]
4. Calculate [tex]\( X^2 \)[/tex]:
[tex]\[ X^2 \approx 1.701 \][/tex]
5. Finally, take the square root of both sides to find [tex]\( X \)[/tex]:
[tex]\[ X \approx \sqrt{1.701} \approx 1.304 \][/tex]
Therefore, the spring must stretch approximately 1.304 meters to store 360 Joules of elastic potential energy.
