An object is suspended from spring 1, and the springs elongation (the distance it stretches) is...

Question:

An object is suspended from spring 1, and the springs elongation (the distance it stretches) is X1. Then the same object is removed from the first spring and suspended from a second spring. The elongation of spring 2 is X2 .X2 is greater that X1.

1) On the same axes, sketch the graphs of the mass versus elongation for both springs.

2) Is the origin included in the graph? Why or why not?

3) Which slope is steeper?

4) At a given mass X2 = 1.6X1. If X2 = 5.3 cm, what is X1?

The Force Constant of a Spring:

The spring is an elastic device and it obeys Hooke's Law. The restoring force of a spring is directly proportional to its elongation. Let the restoring force of a spring be {eq}F {/eq} when its elongation is {eq}x {/eq}.

Then, by Hooke's Law, we can write:

{eq}\displaystyle{F = -kx \ \ \ \ \ \ \ \ \ \ \ \ (\text{-ve sign indicates that the restoring force is just opposite in direction to the elongation})} {/eq}

{eq}k {/eq} is the proportionality constant and this constant is known as the force constant of the spring. The force constant of a spring is defined as the restoring force that the spring exerts when it has a unit elongation. The SI unit of the force constant is {eq}\rm N/m {/eq}.

Answer and Explanation:

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Given


  • The mass of the object: {eq}m {/eq}
  • The elongation of spring 1: {eq}x_1 {/eq}
  • The elongation of spring 2: {eq}x_2 {/eq}


Part (1)


M...

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

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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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