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Work of 3 Joules is done in stretching a spring from its natural length to 11 cm beyond its...

Question:

Work of {eq}3 \ Joules {/eq} is done in stretching a spring from its natural length to {eq}11 \ cm {/eq} beyond its natural length. What is the force that holds the spring stretched at the same distance ({eq}11 \ cm {/eq})?

Find Force And Spring Constant:


The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring for small distances. The force exerted back by the spring is known as Hooke's law.

We use formula here

\displaystyle Force(F)=Spring Constant (k) Distance (x)

Answer and Explanation: 1

Find Force and spring constant

{eq}\displaystyle Force(F)= Spring Constant (k). Distance (x)\\ \text{Find spring constant (k)}\\ W=\frac{1}{2}kx^2\\ k=\frac{2W}{x^2}\\ k=\frac{(600)(100)}{121}\\ \text{Find force (F)}\\ F={spring constant (k)}{distance (x)}\\ F={\frac{(600)(100)}{121}}{\frac{11}{100}}\\ F=54.54 {\text{Newton}}\\ {/eq}


Learn more about this topic:

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

from

Chapter 4 / Lesson 19
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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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