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An archer pulls her bowstring back 0.390 m by exerting a force that increases uniformly from zero...

Question:

An archer pulls her bowstring back {eq}0.390 \ m {/eq} by exerting a force that increases uniformly from zero to {eq}217 \ N {/eq}.

(a) What is the equivalent spring constant of the bow?

(b) How much work does the archer do in pulling the bow?

Hooke's Law:

When the spring is deformed, then the force of spring tends to bring the spring back to the undeformed position. The spring constant of the spring describes how much force is required to deform the spring. Up to the elastic limit, the spring recovers its original shape and size. Mathematically:

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

where:

  • {eq}k {/eq} is the spring constant of elastic material
  • {eq}x {/eq} is the deformation of the elastic material

Answer and Explanation:

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Given data:

  • {eq}x=\rm 0.390 \ m {/eq} is the elongation of the bowstring
  • {eq}F=\rm 217 \ N {/eq} is the applied force
  • {eq}k {/eq} is the spring...

See full answer below.


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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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