Homework Help: physics

Posted by Chris Adison on Wednesday, April 23, 2008 at 1:58pm.

A person wants to participate in a bunjee jump. One end of an elastic band 10m long with a spring constant of 50N/m will be attatched to a basket on a crane and the other end to the waist of the person with a mass of 86kg. The person will step off the edge of the basket to be slowed and brought back by the elastic band before hitting the ground (hopefully 2m above) How high should the crane be? (hint: conservation of energy).

physics - bobpursley, Wednesday, April 23, 2008 at 2:48pm
Initial PE= Final PE
mgh=mg2+ 1/2 k(h-12)^2
check my thinking.
- physics - Chris Adison, Wednesday, April 23, 2008 at 3:54pm
  could i solve it like this?
  Fg=Fs
  mg=kx
  x=mg/k
  x=stretch-Eq value
  stretch=x+Eq value
  than take the stretch value and add 2.
- physics - Chris Adison, Wednesday, April 23, 2008 at 4:29pm
  so when i plug the givens into your equation it should look like this
  
  (86)(9.8)(?)=(86)(9.8)1/2(50)(h-12)^2
  
  Im confused
physics - bobpursley, Wednesday, April 23, 2008 at 4:13pm
No, the system is accelerating, Fg does not equal Fs except at equilibrium.
- physics - Chris Adison, Wednesday, April 23, 2008 at 4:30pm
  so when i plug the givens into your equation it should look like this
  
  (86)(9.8)(?)=(86)(9.8)1/2(50)(h-12)^2
  
  im confused

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