A 0.50 kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from...

Question:

A 0.50 kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 m to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 6.1 to 4.4 m/s. What is the spring constant of the spring?

Spring Constant:

In physics, the spring constant is described as the characteristic of any specific spring, which is equivalent to the dfraction among the force exerting on that spring and its displacement. The spring constant rises with a rise in spring's stiffness.

Answer and Explanation: 1


Given Data:

  • The mass of sphere is {eq}m = 0.50\;{\rm{kg}}{/eq}.
  • The initial stretched length is {eq}h = 0.12\;{\rm{m}}{/eq}.
  • The final stretched length is {eq}h' = 0.23\;{\rm{m}}{/eq}.
  • The initial speed of sphere is {eq}v = 6.1\;{\rm{m/s}}{/eq}.
  • The final speed of sphere is {eq}v' = 4.4\;{\rm{m/s}}{/eq}.


The expression from conservation of energy is given by,

{eq}\begin{align*} {E_f}& = {E_0}\\ K + U + {E_s} &= K' + U' + {E_s}'\\ \left( {\dfrac{1}{2}m{v^2}} \right) + \left( {mgh} \right) + \left( {\dfrac{1}{2}k{h^2}} \right) &= \left( {\dfrac{1}{2}mv{'^2}} \right) + \left( {mgh'} \right) + \left( {\dfrac{1}{2}kh{'^2}} \right) \end{align*}{/eq}

Here, g is the gravitational constant and k is the force constant.


On rearranging the above expression.

{eq}\begin{align*} \dfrac{1}{2}k\left( {h{'^2} - {h^2}} \right) &= \dfrac{1}{2}m\left( {{v^2} - v{'^2}} \right) + mg\left( {h' - h} \right)\\ k &= \dfrac{{m\left( {{v^2} - v{'^2}} \right) + 2mg\left( {h' - h} \right)}}{{\left( {h{'^2} - {h^2}} \right)}} \end{align*}{/eq}


Substitute the values in the above expression.

{eq}\begin{align*} k& = \dfrac{{\left( {0.50\;{\rm{kg}}} \right)\left( {{{\left( {6.1\;{\rm{m/s}}} \right)}^2} - {{\left( {4.4\;{\rm{m/s}}} \right)}^2}} \right) + 2\left( {0.50\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {0.23\;{\rm{m}} - 0.12\;{\rm{m}}} \right)}}{{\left( {{{\left( {0.23\;{\rm{m}}} \right)}^2} - {{\left( {0.12\;{\rm{m}}} \right)}^2}} \right)}}\\ k &= \left( {\dfrac{{8.925 + 1.079}}{{0.0385}}} \right)\;{\rm{N/m}}\\ k &= 259.84\;{\rm{N/m}} \end{align*}{/eq}


Thus, the force constant is {eq}k = 259.84\;{\rm{N/m}}{/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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