We appreciate your visit to Un resorte que funciona como amortiguador de un auto tiene una constante de fuerza tex K 423 25 text N m tex Cuánto debe estirarse. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
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.
Thanks for taking the time to read Un resorte que funciona como amortiguador de un auto tiene una constante de fuerza tex K 423 25 text N m tex Cuánto debe estirarse. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
