Question

Solve for spring constant?

Answer 1

The question involves a 61 kg bungee jumper. The massless bungee cord is normally 25 m long and acts just like a spring when stretched beyond that length. The jumper reaches his lowest point 42.9 m below the bridge

Answer 2

k is supposed to be 160N/m but i dont see how

Answer 3

F = -km is our formula for a spring konstant

Answer 4

F = -k * x
x = 42.9 - 25 = 17.9 m
F = mg = 61 * 9.8 N = 597.8
-597.8 = -k * 17.9
k = ?

Answer 5

I did that and I got 33.39, but I was given the answers as a guide and the answer is supposed to be 160...

Answer 6

can you post a pic of the guide?

Answer 7

sure but its nothing but the answer

Answer 8

not much of a guide then .... pic of the problem and the guide?

Answer 9

ok

Answer 10

does the initial force matter? or velocity?

Answer 11

falling 25 meters builds up steam right?

Answer 12

converting to say kinetic energy may prove useful, just a thought

Answer 13

I working on energy conservation right now.
\[W_{NC}+U_{i}+K_{i}=U_{f}+K_{f}\]

Answer 14

\[
\frac 12 mu^2 - mg\Delta h = \frac 12 kx^2
\]

Answer 15

im not use to those formulas, but im pretty sure this isnt just a simple spring stretch

i might be able to calculate the acceleration involved in order to back track it

Answer 16

or aum seems to be better equiped formulaicly :)

Answer 17

Find the downward velocity 'u' when h = 25m
delta h = x = 17.9 m
m = 61 kg

Answer 18

so wait for that formula I would plug in 17.9 for both x and delta h?

Answer 19

Oh, forget the velocity.
mgh = 1/2 * k * x^2

Answer 20

h = 42.9 m
x = 17.9 m

Answer 21

mgh is the difference in potential energy when he is on the bridge and when he is at the lowest point of the bungee jump.
This lost potential energy is stored in the elastic energy of the bungee cord which is 1/2 * k * x^2 where x is the amount stretched.

It does give 160.08 N / m

Answer 22

so like this then
\[(61)(9.8)(42.9)=\frac{1}{2}k(17.9)^{2}\]

Answer 23

\[
\int kxdx = \frac 12 kx^2 + c
\]

Answer 24

\[k=\frac{2((61)(9.8)(42.9))}{17.9^{2}}=160.08\]

Answer 25

N / m

Answer 26

thank you =)

Answer 27

You are welcome.

Answer 28

-4.9 t^2 + 25 = 0, when t = 2.26

velocity is 9.8*2.26 = 22.15

total energy at start of spring is therefore: 61/2 * (22.15)^2 = 14964 give or take

that disipates in potential energy over 17.9 meters

Answer 29

yall got it :)

Answer 30

for the second part of the question how would I find the magnitude of the acceleration I know that the acceleration is in the y direction but how do I find the magnitude?

Answer 31

(kx - mg) = ma
a = k/m * x - g = 160.08 / 61 * 17.9 - 9.8 = 46.97 - 9.8 = 37.17 m / s^2

Answer 32

Do you have the answer for part b) ?

Answer 33

37.2m/s^2 perfect! =)

Answer 34

Alright :)