A block is dropped onto a spring with k = 34 N/m. The block has a speed of 2.9 m/s just before it...

Question:

A block is dropped onto a spring with {eq}k = 34\ N/m {/eq}. The block has a speed of {eq}2.9\ m/s {/eq} just before it strikes the spring. If the spring compresses an amount of {eq}0.11\ m {/eq} before bringing the block to rest, what is the mass of the block?

Spring:


Spring is an elastic device that is used to absorbed sudden shocks (falling of material on the spring). When an object strikes or falls on the spring, the spring compresses from its unstretched position and stores elastic potential energy in it.

Answer and Explanation: 1


We are given the following data:

  • Mass of the block , {eq}M {/eq},_____
  • Spring constant, {eq}k=34\ \text{N/m} {/eq}
  • Compression of spring, {eq}X=0.11\ \text{m}{/eq}
  • Speed of the block,{eq}V=2.9\ \text{m/s} {/eq}

When the block is dropped on the spring, the kinetic energy stored in the block is expressed by the following equation:

{eq}\begin{align} \text{K.E}&=\dfrac{1}{2}MV^{2}\\[0.3 cm] &=\dfrac{1}{2}\times m\times(2.9\ \text{m/s})^{2}\\[0.3 cm] &=4.205M\ (\text{m}^{2}/\text{s}^{2}) \end{align} {/eq}


When the block encounters an upstretched spring, the kinetic energy of the block is converted into the strain energy of the spring. Calculating the mass of the block by using the following equation:

{eq}\begin{align} \text{K.E}&=\text{Strain energy of spring}\\[0.3 cm] 4.205M\ (\text{m}^{2}/\text{s}^{2})&=\dfrac{1}{2}kx^{2}\\[0.3 cm] 4.205M\ (\text{m}^{2}/\text{s}^{2})&=\dfrac{1}{2}\times (34\ \text{N/m})\times (0.11\ \text{m})^{2}\\[0.3 cm] M&=\boxed{0.0489\ \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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