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When a 9.09 kg mass is placed on top of a vertical spring, thespring compresses 4.30 cm....

Question:

When a 9.09 kg mass is placed on top of a vertical spring, thespring compresses 4.30 cm. Calculate the force constant of thespring.

Spring:

In physics, the spring can regain its original shape after the removal of the applied load. The relation between the spring force and deflection can be obtained from Hook's law. Sping has wide applications in the field of mechanical engineering.

Answer and Explanation: 1


Given Data

  • The magnitude of the mass is: {eq}m = 9.09\;{\rm{kg}} {/eq}.
  • The compression of the spring is: {eq}x = 4.30\;{\rm{cm}} = 0.043\;{\rm{m}} {/eq}.


The spring force will be equal to the weight of the mass, so from Hook's law, we get

{eq}\begin{align*} F &= Kx\\ w &= kx\\ mg &= kx\\ K &= \dfrac{{mg}}{x} \end{align*} {/eq}

Here,

  • {eq}K {/eq} is the force constant.
  • {eq}g {/eq} is the acceleration due to gravity.
  • {eq}w {/eq} is the weight.
  • {eq}F {/eq} is the spring force.

Substitute the values into the above equation

{eq}\begin{align*} K &= \dfrac{{9.09 \times 9.8}}{{0.043}}\\ K &= 2071.67\;{\rm{N/m}} \end{align*} {/eq}


Thus, the force constant of the spring is {eq}\mathbf{2071.67\;{\rm{N/m}}}{/eq}.


Learn more about this topic:

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Practice Applying Spring Constant Formulas

from

Chapter 17 / Lesson 11
3.4K

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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