How much do the bed springs stretch when a person jumps on them?

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The discussion revolves around calculating the stretch of 20 vertical springs in an innerspring mattress when a 48 kg person jumps onto it from a height of 1.95 m. The relevant physics concepts include the conservation of mechanical energy and the spring equation, specifically using the formula 1/2kx². Participants express confusion about setting up the conservation of energy equation and whether to adjust the spring constant for the number of springs involved. There is uncertainty about solving for the stretch (y) algebraically, with some considering the quadratic formula. The thread highlights the need for clarity in applying the spring constant and energy conservation principles to arrive at the correct solution.
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Homework Statement


The innerspring mattress on your grandmother’s bed is held up by 20 vertical springs, each having a spring constant of 5000 N/m. A 48 kg person jumps from a 1.95 m platform onto the innersprings.
The acceleration of gravity is 9.8 m/s2 .
Assume: The springs were initially un- stretched and that they stretch equally (typical old fashioned bed).
Determine the stretch of each of the springs. Answer in units of m.


Homework Equations



consevation of mechanical energy using spring equation, 1/2kx2


The Attempt at a Solution



im not really sure how to setup the conservation of energy equation... if i have it right here then i should solve for y, but i don't see how to do that algebraically. i will always end up with a y on either side. unless I'm supposed to apply the quadratic formula... nope, still a wrong answer. am i setting up the conservation of energy equation right?

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should the spring constant be multiplied by 20 to compensate for the 20 springs, or should the answer be divided by 20?

i don't see where i went wrong in my math...
 
edited
 
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