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

Question

4 months ago

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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?

Physics

2 answers:

kicyunya [3.2K] · Answer 1 · 2026-02-21T22:16:02+03:00

7.2 × 10⁻³ J of energy is required for the pen to extend.

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Further Explanation

Let’s revisit the Elastic Potential Energy formula:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

Ep = elastic potential energy ( J )

k = spring constant ( N/m )

x = amount of spring extension (compression) ( m )

Now, let's solve 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

total compression = x₂ = 1.0 + 6.0 = 7.0 mm

What we need to find:

energy required to extend the pen = ΔEp =?

Solution:

$\Delta Ep = Ep_2 - Ep_1$

$\Delta Ep = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2$

$\Delta Ep = \frac{1}{2}k ( x_2^2 - x_1^2 )$

$\Delta Ep = \frac{1}{2} \times 300 \times [ (7 \times 10^{-3})^2 - (1 \times 10^{-3})^2 ]$

$\Delta Ep = 7.2 \times 10^{-3} \texttt{ Joule}$

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Conclusion:

7.2 × 10⁻³ J is necessary to extend the pen.

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Learn more

Kinetic Energy:

Acceleration:

Car Speed:
Young's Modulus:
Simple Harmonic Motion:

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Answer Details

Grade: High School

Subject: Physics

Chapter: Elasticity

Sav [3.1K] · Answer 2 · 2026-02-21T22:16:03+03:00

Answer:

The pen requires 7.2 mJ of energy to extend.

Explanation:

Provided:

Length = 1.8 cm

Spring constant = 300 N/m

Initial compression = 1.0 mm

Additional compression = 6.0 mm

Total compression = 1.0 + 6.0 = 7.0 mm

We need to determine the energy needed

This energy is equivalent to the variation in spring potential energy

$E=PE_{2}-PE_{1}$

$E=\dfrac{1}{2}kx_{2}^2-\dfrac{1}{2}kx_{1}^2$

Substitute the values into the formula

$E=\dfrac{1}{2}\times300\times(7.0\times10^{-3})^2-\dfrac{1}{2}\times300\times(1.0\times10^{-3})^2$

$E=0.0072\ J$

$E=7.2\ mJ$

Therefore, a total of 7.2 mJ is needed to extend the pen.