How Does Spring Compression Relate to Energy Changes?

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The discussion focuses on the relationship between spring compression and energy changes. The calculation presented for energy loss is clarified as being 300J - 140J, indicating that 140J is the energy retained at maximum compression. The relationship between energy and compression is correctly identified as quadratic, not exponential, with energy increasing proportional to the square of the compression distance. The term "inefficiency" is introduced to describe the energy lost during compression. The final consensus confirms that the retained energy is indeed 140J.
hello478
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Homework Statement
what is percentage efficiency of transfer of KE from child to PE in spring
Relevant Equations
energy equations
part d- ii and iii
ii) my answer is
300-140/300 *100
ke at y = 300
and spring energy at max compression is 140

iii) e is directly proportional to x^2
so it increases exponentially
is my explanation correct?


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hello478 said:
ii) my answer is
300-140/300 *100
I assume you mean (300-140)/300*100%. But 300J - 140J is the energy lost, not the energy retained.
hello478 said:
iii) e is directly proportional to x^2
so it increases exponentially
is my explanation correct?
That's quadratic, not exponential. Exponential would be ##E\propto e^x##
 
haruspex said:
I assume you mean (300-140)/300*100%. But 300J - 140J is the energy lost, not the energy retained.

That's quadratic, not exponential. Exponential would be ##E\propto e^x##
so then what would be the energy retained? only 140??

oh yeah, sorry i forgot, i meant quadratic
 
hello478 said:
so then what would be the energy retained? only 140??
Yes. What you calculated could be called the inefficiency.
 
haruspex said:
Yes.
thank you!!!
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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