Answer :
Answer:
The energy required to extend the pen is 7.2 mJ.
Explanation:
Given that,
Length = 1.8 cm
Spring constant = 300 N/m
Spring is compressed = 1.0 mm
Again, Compressed = 6.0 mm
Total compressed = 1.0+6.0 = 7.0 mm
We need to calculate the required energy
The energy required is equal to the change in spring potential energy
[tex]E=PE_{2}-PE_{1}[/tex]
[tex]E=\dfrac{1}{2}kx_{2}^2-\dfrac{1}{2}kx_{1}^2[/tex]
Put the value into the formula
[tex]E=\dfrac{1}{2}\times300\times(7.0\times10^{-3})^2-\dfrac{1}{2}\times300\times(1.0\times10^{-3})^2[/tex]
[tex]E=0.0072\ J[/tex]
[tex]E=7.2\ mJ[/tex]
Hence, The energy required to extend the pen is 7.2 mJ.
7.2 × 10⁻³ J of energy is required to extend the pen
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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:
length of spring = L = 1.8 cm
spring constant = k = 300 N/m
initial compression = x₁ = 1.0 mm = 1
final compression = x₂ = 1.0 + 6.0 = 7.0 mm
Asked:
energy required to extend the pen = ΔEp = ?
Solution:
[tex]\Delta Ep = Ep_2 - Ep_1[/tex]
[tex]\Delta Ep = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2[/tex]
[tex]\Delta Ep = \frac{1}{2}k ( x_2^2 - x_1^2 )[/tex]
[tex]\Delta Ep = \frac{1}{2} \times 300 \times [ (7 \times 10^{-3})^2 - (1 \times 10^{-3})^2 ][/tex]
[tex]\Delta Ep = 7.2 \times 10^{-3} \texttt{ Joule}[/tex]
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Conclusion :
7.2 × 10⁻³ J of energy is required to extend the pen
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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
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Answer details
Grade: High School
Subject: Physics
Chapter: Elasticity
