Energy help!!!

[natasha.tung]

Hi peoples,

A 60kg bungee jumper falls from bridge 50m above a deep river. The length of the bungee cord when it is not under tension is 30m calculate:

a.) The kinetic energy of the bungee jumper at the instant that the cord begins to stretch beyond its natural length.

b.) The strain energy of the bungee cord at the instand that the tip of the jumper's head touches the water. (her head just makes contact with the water before she is pulled upwards by the cord.) The height of the bungee jumper is 170cm.
Thanks in advance

[Whatlol]

Bascially as he falls, his potential energy will be converted to kintetic energy. So Ug= Ek

mgh= eK

60x9.8x30 = 17640 J

or alternatively

v^2= u^2 +2ax
v ^2= 0 + 2x9.8 x 30
v=24.25ms^-1

Ek= 1/2mv^2
Ek = 17640

[onur369]

I think VCAA takes g as 10ms^-1.
It eases calculations.

[natasha.tung]

Does anyone know how to do b?? I've done a but I just can't seem to get b

[Whatlol]

whats the answer?

[madoscar65]

The answer is 2.9 x 10^4 J

[Killerkob]

Us = kx^(2)/2

'x' is how much the rope is stretched from it's natural position (30m) but we also need to take into account the height of the bungee jumper as they are not part of the rope but are part of the distance from the bridge.
Therefore
x=50-30-1.7
x=18.3m

We can find 'k' using Hooke's law (F=-kx)
k=-F/x
The only force acting on the jumper is gravity.
F=ma
  =60*9.8
F=588N
Therefore
k=-588/18.3Nm^-1 (The negative is negligible for this scenario)

Plugging this into the original equation we get
Us=(588/18.3)(18.3)^(2)/2
Us=2690.1J
Us=2.7kJ

I seriously hope that's correct. It's been too long since I've done physics >.< Also, seriously sorry about my lack of latex knowledge.

[schnappy]

I remember this, you jump straight into strain energy and hookes law... just use energy conservation.

Ek = 0, Eg.p. = 0

ie. All the energy in the system is strain energy.

[filoballerx]

a.) Egp=mgh

therefore as the jumper falls down it loses gravitational potential energy but gains kinetic energy

it says find the kinetic energy of the bungee jumper at the instant that the cord begins to stretch beyond its natural length.

instant that the cord begins which means before it stretches so its 30m

 before the jumper jumps the man have
Egp = mgh           
      = 60x10x50     
      = 30000J

when the jumper jumps the height of the jumper will be 50m - 30m(because the height minus from the bungee cord = 20m

therefore the Egp of the man before the cord stretches is
Egp = 60x10x20
      = 12000 J

it clearly shows that the man loses Egp when it falls down there fore as it loses Egp it gains kinetic energy
there Ek= Egp final - Egp initial
        Ek= 30000 - 12000
           = 18000J is your kinetic energy

sorry if it doesnt make sense, im not really good at teaching hehehe