Asked by Katie
The innerspring mattress on your grand- mother’s bed is held up by 23 vertical springs, each having a spring constant of 6000 N/m. A 43 kg person jumps from a 1.93 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 (typi- cal old fashioned bed).
Determine the stretch of each of the springs. Answer in units of m
The acceleration of gravity is 9.8 m/s2 .
Assume: The springs were initially un- stretched and that they stretch equally (typi- cal old fashioned bed).
Determine the stretch of each of the springs. Answer in units of m
Answers
Answered by
Elena
PE-> KE -> PE(springs)
mgh =23 kx²/2
x=sqrt{2mgh/23k} =
=sqrt{2•43•9.8•1.93/23•6000} =0.012 m
mgh =23 kx²/2
x=sqrt{2mgh/23k} =
=sqrt{2•43•9.8•1.93/23•6000} =0.012 m
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