PHYSICS HELP PLEASE!!!!!...?

Question

A bungee jump of mass 60 kg jumps from a bridge tied to an elastic rope which becomes taut after he falls 10m. Consider the jumper when he has fallen another 10 m and is travelling at 15 m/s.

State a) a form of energy that has been lost, b) two forms of energy that have been gained

c) work out how much energy is stored in the rope. take g=10m/s2 and ignore air resistance..

Please show answer with the steps n not just the answer. When it says that the bungee jumper has fallen 10 m more...does it mean that the rope breaks?

I need it now please!......

Answer 1

The bungee works the same as a spring, i.e. energy stored in the bungee is just 1/2 * K * X^2.  The problem is that you are not given the "bungee" constant K, but that's OK you don't need to figure it out (although this problem would be a lot easier if you knew it...).  First you need to assign some relative locations.  Call to top (where he jumps) state "0", where the bungee becomes taunt state "1", and the bottom (where V=15 m/s) state "2".

At state "o" all his energy is in gravitational potential energy.  There's no energy stored in the bungee (it's not stretched yet) and there's no kinetic energy (he's not moving).  If we use state "2" as the zero point for gravitational potential (this is always an arbitrary ref point), then his TOTAL energy at state "0" is just m * g * h and is constant FOR ALL OTHER STATES (conservation of energy principle). So at state "0" his total energy is just::

PE = m * g * h = 60 * 10 * 20 = 12000 joules

This total system energy is constant for states "1" and "2", it is just converted to other forms of energy.  

For state "1" there is still no energy stored in the bungee (it just became taunt, not yet stretched), there is still some PE, but less, and some KE since he's now moving:

12000 J = PE + KE = m * g * h + 1/2 * m * V^2

you'll need to find the guy's velocity after freefalling 10 m:

V1^2 = Vo^2 + 2 * g * h

V1^2 = 0 + 2 * 10 * 10

V1^2 = 200

V1 = 14.1 m/s

If you plug this V1 back into the energy equation above you get pretty close to the 12000 J, at least within rounding error.  OK on to state "2"...

At state "2" we assume PE = 0 (this was our ref point for gravity, so all the energy has been converted to kinetic (KE) and "stored bungee" (BE) energy:

12000 J = KE + BE

You're given the velocity V2 = 15 m/s, so KE2 is just

KE2 = 1/2 * g * V2^2

KE2 = 1/2 * 10 * 15^2 = 1125 J; therefore

12000 J = 1125 J + BE, so energy stored in the bungee at state "2" is

12000 - 1125 = BE = 10875 J

Seems like a lot, but consider if he weren't attached to the bungee his freefall velocity after 20 m would be

V^2 = 2 * g * d = 20 m/s instead of 15 m/s...

So after all of that, the problem asks for the following:

a) Energy lost = gravitational potential energy

b) Energy gained = kinetic & "spring" or bungee stored energy

c) Energy stored in bungee @ state "2" = 10875 J

Good Luck!