Physics help
A 66kg trampoline artist hits a trampoline with a velocity of 7.5m/s. If the trampoline acts as a spring with constant 6.5*10^4N/m, what is the distance he depresses the trampoline.
My reasoning:
Calculate KE=1/2mv^2=1856.25J
KE=W=Fd
F=kx
x=d (not sure of this)
==>KE=kx^2
1856.25=(6.5*10^4)*x^2
x=0.17m
Says it's wrong. Probably because x=/=d or KE=/=Fd.
September 26, 2013
2 Comments • Newest first
[quote=HolyDragon]Here's my guess.
The total energy of the gymnist at that time is all Kinetic energy (1856.25J)
That's the energy applied on the trampoline.
Spring energy formula
PE=1/2k x^2
sub it in and it should work.[/quote]
It worked. Thanks
Here's my guess.
The total energy of the gymnist at that time is all Kinetic energy (1856.25J)
That's the energy applied on the trampoline.
Spring energy formula
PE=1/2k x^2
sub it in and it should work.