Asked by Michael
A spring-like trampoline dips down 0.06 m when a particular person stands on it. If this person jumps up to a height of 0.29 m above the top of the trampoline, how far with the trampoline compress when the person lands? Include units.
This is a question on a review guide that we are doing in my physics class and I have no clue how to begin...
This is a question on a review guide that we are doing in my physics class and I have no clue how to begin...
Answers
Answered by
bobpursley
The compressed trampoline :
energy in the compression=1/2 k (.06^2)
added energy from jump=M*g*.29
but at rest, initially, force=k(.06)=Mg
so final energy= 1/2 k x^2=1/2 k(.06+.29)^2 (assuming he jumped from the initial depressed position)
solve for x
energy in the compression=1/2 k (.06^2)
added energy from jump=M*g*.29
but at rest, initially, force=k(.06)=Mg
so final energy= 1/2 k x^2=1/2 k(.06+.29)^2 (assuming he jumped from the initial depressed position)
solve for x
Answered by
R_scott
the energy stored in the trampoline is the gravitational energy
... 1/2 k x^2 = m g h ... k = 2 * m * g * .06 / .06^2 = 100/3 m g
in the 2nd case , there is more gravitational energy
... m , g , and k are the same ... x and h change
1/2 k x^2 = m g (x + .29) ... x^2 = .06 (x + .29)
... 1/2 k x^2 = m g h ... k = 2 * m * g * .06 / .06^2 = 100/3 m g
in the 2nd case , there is more gravitational energy
... m , g , and k are the same ... x and h change
1/2 k x^2 = m g (x + .29) ... x^2 = .06 (x + .29)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.