The work required to strecth the spring is about 2.0 Joules
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Further explanation
Let's recall Elastic Potential Energy formula as follows:
[tex]\boxed{E_p = \frac{1}{2}k x^2}[/tex]
where:
Ep = elastic potential energy ( J )
k = spring constant ( N/m )
x = spring extension ( compression ) ( m )
Let us now tackle the problem!
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Given:
spring constant = k = 2500 N/m
extension = x = 4.00 cm = 4.00 × 10⁻² m
Asked:
work required = W = ?
Solution:
We will calculate the work required to stretch the spring as follows:
[tex]W = E_p[/tex]
[tex]W = \frac{1}{2} k x^2[/tex]
[tex]W = \frac{1}{2} \times 2500 \times ( 4.00 \times 10^{-2} )^2[/tex]
[tex]W = 1250 \times ( 16.00 \times 10^{-4} )[/tex]
[tex]W = 20000 \times 10^{-4}[/tex]
[tex]W = 2.0 \times 10^4 \times 10^{-4}[/tex]
[tex]W = 2.0 \texttt{ J}[/tex]
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Conclusion:
The work required to strecth the spring is about 2.0 Joules
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Learn more
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
- Young Modulus : https://brainly.com/question/9202964
- Simple Harmonic Motion : https://brainly.com/question/12069840
- Light Ideal Spring Speed : https://brainly.com/question/13050880
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Answer details
Grade: High School
Subject: Physics
Chapter: Elasticity
