so in this problem you have a man bungee-jumping from a hot airballon 65 m high.
ID: 1764609 • Letter: S
Question
so in this problem you have a man bungee-jumping from a hot airballon 65 m high. He wants to stop 10 meter below the ground(assume cord weight is negligible)and the system obeys hook law. Hedoes a test with the same type of cord 5m long in which he allowshis body weight to stretch it from rest which extends the cord 1.50m. He intends to drop from the rest at the point where the top endof a longer section of the cord is attached to the stationary hotair ballon. It asks what length of cord should he use? and whatmaximun acceleration will he experienceI know that with hooks law wher Fs = -kx
in which you have to solve back for k
Fs = -1.5k
So i knew that this force was equivalent to the normal forcefor which the mgy is compared to
mgy = -1.5k
So i plugged in values to try to solve back for the value of m
(9.8)(m)(1.5) = -1.5k
which you can divide out the 1.5 k
and h ave k = -9.8m
from here wasnt too sure what to do
Explanation / Answer
(a) from the given for the 5 m cord the springconstant is given by F = k x F = m g m g = k (1.50 m) for a longer cord of length L the strech distancewill be longer so we get k = (5 m / L) (m g / 1.50 m) = .... (m g / L) from teh conservation principle we get (KE + PEg +PEs)i = (KE + PEg +PEs)f 0 + m g yi + 0 = 0 + m gyf + (1 / 2) k xf2 m g (yi - yf) = (1 / 2) kxf2 = (1 / 2) ........(m g / L)xf2 yi - yf = 65 m = L + xf 65.0 m L = (1 / 2) ..... (65.0 m - L)2 solve for L L = ........ m (b) we got k = .... (m g / L) xmax = xf =65.0 m - L =......... m as F = m a + k xmax - m g = m a solve for acceleration a a = ......... m / s2 solve for L L = ........ m (b) we got k = .... (m g / L) xmax = xf =65.0 m - L =......... m as F = m a + k xmax - m g = m a solve for acceleration a a = ......... m / s2
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