# Homework Help: Gr12 Energy Prob (Bungee)

1. Dec 20, 2008

### Uncle6

1. The problem statement, all variables and given/known data

"A bungee jumper of mass 64.5 kg ( including safety gear) is standing on a platform 48.0 m above a river. The length of the unstretched bungee cord is 10.1 m. The force constant of the cord is 65.5 N/m. The jumper falls from rest and just touches the water at a speed of zero. The cord acts like an ideal spring. Use conservation of energy to determine the jumper's speed at a height of 12.5 m above the water on the first fall."

m=64.5 kg
fconstant= 65.5 N/m
dT=48.0m
v3=0 m/s
unstreched bungee cord = 10.1 m
y2 = 12.5m
RTF = v2

2. Relevant equations
W=0
Et1=Et2

3. The attempt at a solution

I drew a diagram first of total distance 48.0 m. There is 10.1 down from the top and 12.5 m up from the bottom.
I make x1 the distance that remains
So: x1 = 48.0-10.1-12.5 = 37.9m
x2= x1+12.5 = 25.4

0 = ΔEg+ΔEk+ΔEe
(Rearrange)
v2 = sqrt(2(mgΔy+1/2m(vsub3)^2+1/2k(xsub2)^2 -1/2k(xsub1)^2)/m)
(at bottom, the velocity(vsub3) is 0, I plug in #s)
= 23.6 m/s (books says 6.37 m/s)

2. Dec 20, 2008

### LowlyPion

I think you've almost got it. I get 6.3748 m/s.

The Difference in Potential Energy of gravity minus what gets stored in the cord is what's available for KE.

The 35.5m drop to the 12.m mark is 22,439.55 J.
The PE of the cord is 21,128.99 J
the difference is 1310.56 J which is the KE of the jumper at that point.

Check the signs of your equation.

3. Dec 20, 2008

### Staff: Mentor

Something doesn't seem right about these numbers (or maybe I just need more coffee?). Compare the decrease in gravitational PE when the jumper falls 48 m to the increase in spring PE when he just touches the water (how much has the cord stretched?).

4. Dec 23, 2008

### Uncle6

Im still having trouble with this question. I don't understand the Elastic Pot Energy. Could explain to me the heights you used and why.

I don't understand how to get 35.5 m

5. Dec 23, 2008

### LowlyPion

You start at 48 m.

They want his velocity at 12.5 m. 48 - 12.5 = 35.5 m. That's how much potential energy has been converted from gravity.

Now if there was no bungee attached to his feet he'd be trucking pretty darn quickly toward impact. But that's where the bungee enters. After he has had the enjoyment of 10.1 m of free fall he reaches the relaxed length of his bungee. Now comes the retarding effect. As he drops lower and lower the bungee gets stretched and as it gets stretched that kinetic energy that he was gaining on the way down gets drained away starting at 10.1 m.

Happily they say the bungee will save him before impact. But at the point of interest 12.5 m he still has kinetic energy and hence velocity.

So ... what he has then is the potential energy of the drop to that point propelling him still less what the bungee has been robbing away as it stretches to that point (what went into spring potential energy).

Hence m*g*h -1/2kx2 = 1/2*m*v2

h is the fall to that point. x is how far it's stretched since he passed 10.1m.

6. Dec 25, 2008

### Uncle6

K, got it thanks.