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The stored potential energy in the trampoline at maximum compression is \frac{1}{2} k X^2. It...

Question:

The stored potential energy in the trampoline at maximum compression is {eq}\frac{1}{2} k X^2 {/eq}. It equals the gravitational potential energy at maximum height {eq}= MgH {/eq}. ({eq}\frac{1}{2} k X^2 = MgH {/eq}, as Kinetic energy is zero in both cases). Given, {eq}X = 0.5\ m, g = 9.8\ m/s^2, H = 2.5\ m, M = 115\ kg, k = {/eq} the spring constant. Solve for {eq}k {/eq}.

Elastic Potential Energy:

When an elastic material is subjected to compression or stretching, the potential energy is stored in the material varies linearly with the square of the stretch length. This potential energy will be the same as the gravitational potential energy.

Answer and Explanation:

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Given data

  • The stretch length: {eq}X = 0.5\;{\rm{m}}{/eq}
  • Maximum height: {eq}H = 2.5\;{\rm{m}}{/eq}
  • Mass: {eq}M = 115\;{\rm{kg}}{/eq}

{eq}\\ {/eq}

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Practice Applying Spring Constant Formulas

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Chapter 17 / Lesson 11
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In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.


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