Compressed spring, doubling speed of impact

AI Thread Summary
The discussion revolves around calculating the maximum compression of a spring when a shoe's impact speed is doubled. The initial calculations indicate that the shoe, weighing 0.211 kg and impacting a spring with a constant of 102 N/m, compresses the spring 0.117 m at a speed of 2.08 m/s. When the speed is doubled, the kinetic energy increases fourfold, leading to a new kinetic energy of 1.826 J. The user sets up a quadratic equation to find the new compression but struggles to solve it correctly. The conversation highlights the challenge of finding an alternative method to solve the problem without relying solely on the quadratic equation.
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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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