In: Physics
A bungee jumper of mass 110 kg jumps from a cliff of height 120m. The massless relaxed bungee cord has a length of 15 m. Ignore the height of the body of the jumper. K=125N/m. (a) Find the velocity of the jumper right before the bungee starts to stretch. (b) Find the distance the cord stretches.
Please explain the steps. Thank you!
a)
hi = initial height of jumper at the start of jump at the top of cliff = 120 m
L = relaxed length of the cord = 15 m
hf = final height of jumper as the cord starts to stretch = hi - L = 120 - 15 = 105 m
v = velocity of jumper as the cord starts to stretch
Using conservation of energy
initial potential energy = final potential energy + kinetic energy
mg hi = mg hf + (0.5) m v2
g hi = g hf + (0.5) v2
(9.8 x 120) = (9.8 x 105) + (0.5) v2
v = 17.15 m/s
b)
hi = initial height of jumper at the start of jump at the top of cliff = 120 m
L = relaxed length of the cord = 15 m
x = stretch in the cord
hf = final height of jumper as the cord starts to stretch = hi - (L + x) = 120 - (15 + x) = 105 - x
k = spring constant = 125 N/m
m = 110 kg
Using conservation of energy
initial potential energy = final potential energy + kinetic energy
mg hi = mg hf + (0.5) k x2
(110 x 9.8 x 120)= (110 x 9.8) (105 - x) + (0.5) (125) x2
x = 26.875 m/s