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).

Answer 1

Initial PE= Final PE
mgh=mg2+ 1/2 k(h-12)^2
check my thinking.

Answer 2

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.

Answer 3

No, the system is accelerating, Fg does not equal Fs except at equilibrium.

Answer 4

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

Answer 5

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