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
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...
2 answers
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)
9 answers