A 85 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 3.0 m/s. (Ignore small changes in gravitational potential energy.)
(a) How fast is he going as he lands on the trampoline, 3.0 m below?
= 8.234 m/s
(b) If the trampoline behaves like a spring with spring stiffness constant 5.2 104 N/m, how far does he depress it?
=.........m
- Please help me how to do part b, i already got part a, which is right, but i don't get how to find part b, so please if you could help me, i would really appreciate it, Thanks :)))))
2007-10-13 10:32:24 · 5 answers · asked by Nikita 1 in Science & Mathematics ➔ Physics
SORRY
IT'S SUPPOSE TO BE 5.2 x 10^4 N/m
NOT 5.2 104
2007-10-13 10:45:36 · update #1
You have to include the additional potential energy of the fall through distance x. The energy equality is
kx^2/2 = mv^2/2 + mgx
This results in a quadratic in x:
k/2*x^2 - mg*x - mv^2/2 = 0
Solve it and take the positive root.
The answer is between 0.3 and 0.4 m.
P.S. I checked your 1st answer and I agree.
2007-10-13 12:34:39 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
The answer for part B would be 52m because all you have to do is divide 104 by 2 and you'll get 52. The pressure of the spring will put extra pressure to go that far. Do you get it?
2007-10-13 10:42:30 · answer #2 · answered by Anonymous · 0⤊ 1⤋
Use energy formula.
mgh + (1/2) mv^2 = 1/2 kx^2.
Assuming gravitational potential energy is 0, then :
x= sqrt (mv^2/k)
Good luck.
2007-10-13 10:48:03 · answer #3 · answered by KarenaT 3 · 0⤊ 0⤋
No. He has preliminary and unchanging p.c.. V1. This p.c.. is larger TO the thrown soccer, VF. So the comprehensive VF will continually be bigger than V1 as V1 is integrated in it. till he passes it vertically. yet there could be no person up there to seize it. Dot
2016-11-08 05:32:33 · answer #4 · answered by Anonymous · 0⤊ 0⤋
My guess would be 50.8 n/m
2007-10-13 11:18:50 · answer #5 · answered by San Juanita P 1 · 0⤊ 0⤋