How Much Does the Rope Stretch When a Circus Performer Hangs at Rest?

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SUMMARY

The discussion focuses on calculating the extension of an elastic rope when a 55.0 kg circus performer hangs at rest. The performer oscillates with a period of 2.60 seconds, and the rope follows Hooke's Law. By applying the formula T = 2π√(m/k), the effective spring constant k is determined. At rest, the upward spring force (kx) equals the gravitational force (mg), allowing for the calculation of the rope's extension beyond its unloaded length.

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  • Calculate the effective spring constant k using the period formula T = 2π√(m/k)
  • Determine the gravitational force acting on the performer using F = mg
  • Use the relationship kx = mg to find the extension x of the rope
  • Explore real-world applications of Hooke's Law in engineering and physics
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A 55.0 kg circus performer oscillates up and down at the end of a long elastic rope at a rate of once every 2.60 s. The elastic rope obeys Hooke's Law. By how much is the rope extended beyond its unloaded length when the performer hangs at rest?

Can someone help me to find the answer step by step?
 
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The period of the system (since it obeys Hooke's law) is given by T = 2*pi*sqrt(m/k). You can use the given data to solve for k, the effective spring constant.

When the performer is at rest, the net force on him must be zero. Therefore, the upward spring force, kx, must equal the gravitational force on the person.
 

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