PHYSICS HELPS PLS!!!I've posted this questions twice..no on answers it!?

Question

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

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

Sorry I messed this up the other day; I used "g" in the kinetic energy expression instead of "m".  Here's the whole thing corrected:

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 * m * V2^2

KE2 = 1/2 * 60 * 15^2 = 6750 J; therefore

12000 J = 6750 J + BE, so energy stored in the bungee at state "2" is

12000 - 6750 = BE = 5250 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" = 5250 J

SORRY ABOUT THAT!

Answer 2

a) Gravitational potential

b) Elastic potential, Kinetic

Answer 3

Hi! Give you some hints.

That the elastic rope becomes taut doesn't mean the rope breaks, It means that the rope begins to exert tensile force on the jumper.

Answer 4

a)gravitational potential energy was lost

b)Elastic Potential Energy and Kinetic energy gained

c) The amount of energy stored in the rope is....

EPE=GPE-KE=MGH-1/2 mv^2=60x10x20-0.5x60x15^2

         =5250 J

         =5.25 kJ

no the rope doesn't break; the rope begins to store energy after the first 10m

Answer 5

c) Hints

PE = gravitational potential energy

KE = kinetic energy

EE = elastic energy

For the first 10 m he is free falling.  Show that on a force diagram at t = 0 and t = t1 (at 10 m)

Also draw force diagram at t = t2,  where the bungee chord is an added force.

Note that the total energy is the same at t = 0 and t = t2.  Therefore

PE + KE + EE (at t = 0) = PE + KE + EE (at t = t2)

Now you should have formulas for all of these (if you don't, look them up in your text book)

Careful, now, because EE and KE both equal 0 at t = 0

Answer 6

Maybe no one wants to do your homework for you.

Answer 7

a)Gravitational potential energy is lost because this energy decreases when you come closer to the Earth's surface.

b)Kinetic energy of the person has been gained. And also the elastic potential energy stored in the rope has been gained.

c)Let the gravitational potential energy at the top of the jump be taken as 0.

At the top of the jump: -

Gravitational P.E. = 0

Elastic P.E. in the rope = 0 (because the rope is slack)

K.E. = 0 because the speed = 0

So, total energy = 0---------------(1)

When the jumper has fallen 10 m below the point at which the rope becomes taught, then

height below the start of jump = 10 m + 10 m = 20 m

Gravitational P.E. = - mg*(20 m) = - 60 * 10 * 20 = -12000 J

k.E. = 1/2 mv^2 = 1/2 * 60 * 15^2 = 1/2 * 60 * 225 = 6750 J

Let Ue = elastic P.E. in the rope

Total energy = -12000 + 6750 + Ue

Or total energy = Ue - 5250----------------------(2)

From (1) and (2), by conservation of energy

Ue - 5250 = 0

Or U2 = 5250 J

Ans: 5250 J

The rope does not break. It becomes longer the way a spring becomes longer when it is stretched.

Answer 8

lose gravitational energy

gain kinetic and elastic potential energy

The question means that the mass is in free fall for 10m before the elastic force in the rope kicks in, then it is decelerated for the next 10m and ends up with 15m/s of kinetic energy leftover.

m=60kg

v=15m/s

initially m has h=20m of gravitational energy

finally it has x=10m of elastic potential enregy and 15m/s of kinetic energy

energy is conserved

initial energy = final energy

gravitational = elastic + kinetic

mgh = elastic + ½mv²

elastic = mgh - ½mv² = m(gh - ½v²) = 60(10*20 - ½*15²) = 5250J

stored in the rope

,.,.,