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The spring in a retractable ballpoint pen is 1.8 cm long, with a 300 N/m spring constant. When the pen is retracted, the spring is compressed by 1.0 mm. When you click the button to extend the pen, you compress the spring by an additional 6.0 mm.

How much energy is required to extend the pen?

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.

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Rewritten by : Brahmana

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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Answer details

Grade: High School

Subject: Physics

Chapter: Elasticity