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When a 2.70-kg object is hung vertically on a certain light spring described by Hooke's law, the...

Question:

When a 2.70-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 3.06 cm.

A) What is the force constant of the spring? (in N/m)

B) If the 2.70-kg object is removed, how far will the spring stretch if a 1.35-kg block is hung on it? (in cm)

C) How much work must an external agent do to stretch the same spring 6.40 cm from its unstretched position? (in J)

Work:

When a spring is stretched from one position to another, the work required to stretch the sting is expressed by the relation:

{eq}W=\dfrac{1}{2}kx^{2} {/eq}

W is the work required by spring in {eq}\text{(J)} {/eq}

k is the stiffness of the spring {eq}\text{(N/m)} {/eq}

x is the compression of spring {eq}\text{(m)} {/eq}

Answer and Explanation:

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(a)The force constant of the spring is

  • The mass of the object {eq}m=2.70\ \text{kg} {/eq}
  • Strech of the spring {eq}x=3.06\ \text{cm} {/eq}

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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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