Compressed spring, doubling speed of impact

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SUMMARY

The discussion centers on a physics problem involving a 0.211 kg shoe dropped onto a spring with a spring constant of 102 N/m. When the shoe's speed just before impact is doubled, the kinetic energy increases fourfold, leading to a new kinetic energy of 1.826 J. The participant attempts to find the maximum compression of the spring by equating kinetic energy to net work done, resulting in a quadratic equation: m*g*x - 0.5*k*x^2 = -1.826 J. The participant seeks alternative methods to solve for maximum compression but is encouraged to resolve the quadratic equation correctly.

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  • Understanding of kinetic energy and its relationship with speed
  • Familiarity with Hooke's Law and spring constants
  • Ability to solve quadratic equations
  • Basic knowledge of work-energy principles in physics
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  • Review the derivation of kinetic energy and its implications when speed changes
  • Study Hooke's Law and its application in spring compression problems
  • Practice solving quadratic equations in the context of physics problems
  • Explore alternative methods for solving energy conservation problems in mechanics
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of spring dynamics and energy transformations.

shrutij
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Homework Statement


A 0.211 kg shoe is dropped onto a vertically oriented spring with a spring constant of 102 N/m. The shoe becomes attached to the spring upon contact, and the spring is compressed 0.117 m before coming momentarily to rest.
If the speed of the shoe just before impact is doubled, what is the maximum compression achieved by the spring?


The Attempt at a Solution


Before this question, I figured out the speed the shoe was traveling at just before it hit the spring to be 2.08 m/s, which was correct. I also figured out the work done by the spring force to be -0.698 J and the work done by the weight of the shoe to be 0.242 J.
I know that when speed doubles, K.E. quadruples, which means the net work done must increase by 4 as well.
Wnet= Wgrav+Wspring=0.242 -0.698=-0.456 J.
I found the new K.E. to be 1.826 J (with doubled speed).
When I equate KE to Wnet, I get a quadratic: m*g*x -0.5*k*x^2= -1.826 J (new K.E.)
Is there no other way other than solving this quadratic to get to the new compression?
HELP, please!
 
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shrutij said:

Homework Statement


A 0.211 kg shoe is dropped onto a vertically oriented spring with a spring constant of 102 N/m. The shoe becomes attached to the spring upon contact, and the spring is compressed 0.117 m before coming momentarily to rest.
If the speed of the shoe just before impact is doubled, what is the maximum compression achieved by the spring?

The Attempt at a Solution


Before this question, I figured out the speed the shoe was traveling at just before it hit the spring to be 2.08 m/s, which was correct. I also figured out the work done by the spring force to be -0.698 J and the work done by the weight of the shoe to be 0.242 J.
I know that when speed doubles, K.E. quadruples, which means the net work done must increase by 4 as well.
Wnet= Wgrav+Wspring=0.242 -0.698=-0.456 J.
I found the new K.E. to be 1.826 J (with doubled speed).
When I equate KE to Wnet, I get a quadratic: m*g*x -0.5*k*x^2= -1.826 J (new K.E.)
Is there no other way other than solving this quadratic to get to the new compression?
HELP, please!
There may be some other way, but what's wrong with solving the quadratic equation?
 
I tried solving the quadratic equation, but didn't get the right answer. Can someone tell me where I went wrong with my approach?
thanks
 

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