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Challenging physics question, hooke's law and conservation of energy

  • Thread starter itunescape
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Homework Statement


Obeying hooke's law. A hot air balloon is 65.0m from the ground. The bungee jumper wants to jump with a uniform elastic cord up to 10m above the ground. During a preliminary test, the cord at rest was 5.0m and when he got on it stretched 1.50m. a) What length of cord should he use? b)What maximum acceleration will he experience?

(The problem was much longer than this i swear, but above is all that is given).

Homework Equations


ke(initial) + Pegrav(initial)+Pespring(initial)=Ke(final)+Pegrav(final)+Pespring(final )

1/2mv^2+mgh+1/2kx^2=1/2mv^2+mgh+1/2kx^2
Fs= -kx
F=ma
F=mg

The Attempt at a Solution


The question is long but only distances are given... I made an attempt to solve for k and mass but it was futile. There are too many variable to solve for, unless by some miracle, the unknowns cancel out.
total D= 65.0m
final D= 10.0m
total D-final D= 55.0m total distance the man will bungee jump.
Preliminary test data:
at rest cord = 5.0m
mass stretched the cord 1.50m
new cord length= 7.50m

-The length of the cord must be less than 55 but needs to stretch to 55m. That I can safely assume.
-The Kinetic energy at the initial point must zero
-The potential energy at the final point must be zero bc the cord can only stretch that far (55 m down).

conservation of energy:initial = final

1/2mv^2+mgh+1/2kx^2=1/2mv^2+mgh+1/2kx^2
0 + m*g*h + 1/2k*x^2= 1/2m*v^2 + 0 + 1/2k*x^2
m*g*h+ 1/2k*x^2=1/2m*v^2 + 1/2k*x^2
plug in what i can assume to be the right #s:

m*9.8*65.0 + 1/2 k*x^2= 1/2m*v^2 + 1/2k*10^2

from here i'm stuck because there is m, v, k to solve for along with x.
So i took a desperate approach >>>
from the data given on the preliminary test within the problem, i took a porportional approach.

5.00m X
------ = -------
7.50m 55.0m

X= 36.7 m (This is not the correct answer, but i think it was a nice try lol)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
My peer's approach:
1/2mv^2+mgh+1/2kx^2=1/2mv^2+mgh+1/2kx^2
k=mg/x
v= sqrt(2*g*h)
in conservation of energy equation factor out the mass.
1/2m*(sqrt2*g*h)^2+mgh+1/2(mg/x)*x^2=1/2m*(sqrt2*g*h)^2+mgh+1/2(mg/x)*x^2
1/2m*(sqrt2*9.8*55)^2 +m*9.8*55+1/2(m9.8/1.5)*1.5^2=1/2m*(sqrt2*9.8*10)^2+m*9.8*10+1/2(m*9.8/x)*x^2

he stopped here because the equation got too messy and made no sense.
I am not asking for an answer, just the logic to solving such a complicated problem. What is the first step suppose to be? solve for mass? does that get canceled out in the end?
Thank you for your time for attempting such a lengthy problem.
 

Answers and Replies

  • #2
Shooting Star
Homework Helper
1,977
4
Thank you for your time for attempting such a lengthy problem.
This short problem has been stretched more than the bungee cord... :smile:

The salient point to recognize here is that the effective spring constant of multiple springs or strings connected in series is given by 1/k_eff = ∑(1/ki). For example, if two identical cords are joined in series to make a new one double the length, then the new k' is given by:

1/k' = 1/k + 1/k => k' = k/2. This follows directly from Hooke's law.

Using this, the spring constant k2 of the longer cord should be k2 = (L1/L2)K, where L1 and L2 are lengths of the shorter and longer cords respectively. L2 is the unknown to be found, and k1 of the shorter test cord of length L1 is given.

Now equate the initial PE_grav to the final PE_spring.

I hope that is enough for you proceed further.
 

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