The length of a spring increases by 7.2 cm from its relaxed length when a mass of 1.4 kg is...

Question:

The length of a spring increases by 7.2 cm from its relaxed length when a mass of 1.4 kg is hanging in equilibrium from the spring.

(a) What is the spring constant?

(b) How much elastic potential energy is stored in the spring?

(c) A different mass is suspended and the spring length increases by 12.2 cm from its relaxed length to its new equilibrium position. What is the second mass?

Mass-Spring System:

A spring offers resistance to deformation by a restoring force, which acts opposite to the deformation. When a block is kept hanging at one end of the spring in vertical position, the restoring force of elongated spring balances the weight of the block.

Answer and Explanation: 1

Given Data

  • mass of the hanging block, {eq}m\ = 1.4\ \text{kg} {/eq}
  • elongation in the length of the spring, {eq}\Delta x\ = 7.2\ \text{cm}\ = 7.2\times 10^{-2}\ \text{m} {/eq}

(a) Finding the spring constant (k)

{eq}\begin{align} \text{At equilibrium: Spring Restoring Force = Weight of the block}\\[0.3cm] k\times \Delta x\ &= m\times g\\[0.3cm] k\times 7.2\times 10^{-2}\ \text{m}\ &= 1.4\ \text{kg}\times 9.8\ \text{m/s}^2\\[0.3cm] k\ &= 1.9\times 10^2\ \text{N/m} \end{align} {/eq}

(b) Finding the Potential Energy (U) stored in the spring

{eq}\begin{align} \text{The Potential Energy stored in the spring is calculated as:}\\[0.3cm] U\ &= \dfrac{1}{2}\times k\times (\Delta x)^2\\[0.3cm] U\ &= \dfrac{1}{2}\times 1.9\times 10^2\ \text{N/m}\times (7.2\times 10^{-2}\ \text{m})^2\\[0.3cm] U\ &= 0.49\ \text{J} \end{align} {/eq}

(c) Finding the mass (m' ) of the new block, to get the elongation in the spring as {eq}\Delta x'\ = 12.2\ \text{cm}{/eq}

{eq}\begin{align} \text{Similar to (a), Spring Restoring Force = Weight of the block}\\[0.3cm] k\times \Delta x'\ &= m'\times g\\[0.3cm] 1.9\times 10^2\ \text{N/m}\times 12.2\ \text{cm}\ &= m'\times 9.8\ \text{m/s}^2\\[0.3cm] 1.9\times 10^2\ \text{N/m}\times 0.122\ \text{m}\ &= m'\times 9.8\ \text{m/s}^2\\[0.3cm] m'\ &= 2.37\ \text{kg} \end{align} {/eq}


Learn more about this topic:

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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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