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A biologist hangs a sample of mass 0.725kg on a pair of identical, vertical springs in parallel...

Question:

A biologist hangs a sample of mass 0.725kg on a pair of identical, vertical springs in parallel and slowly lowers the sample to equilibrium, stretching the springs by 0.200m. Calculate the value of the spring constant of one of the springs.

Hooke's Law:

When the mass is attached to the vertical coiled spring, then the spring stretches, until the elastic force of the spring and the gravitational force acting on the mass becomes equal. Under static equilibrium, the net force acting on the attached mass is zero. According to Hooke's law:

{eq}\rm F=kx {/eq}

where:

  • {eq}k {/eq} is the modulus or force constant of the spring
  • {eq}x {/eq} is the stretch produced in the spring

Answer and Explanation:

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Given data:

  • {eq}m=\rm 0.725 \ kg {/eq} is the sample of the mass
  • {eq}k {/eq} is the spring constant
  • {eq}x=\rm 0.200 \ m {/eq} is the elongation...

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Hooke's Law & the Spring Constant: Definition & Equation

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Chapter 4 / Lesson 19
201K

After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.


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