I Need Help with a Physics Problem?

Question

If anyone can help me out with this I would really appreciate it. I can not come up with the right answer that's in the book.

In the dangerous "sport" of bungee jumping, a daring student jumps from a hot-air balloon with a specially designed elastic cord attached to her waist. The unstretched length of the cord is 25.0 m, the student weighs 700 N, and the balloon is 36.0 m above the surface of a river below.

Calculate the required force constant of the cord if the student is to stop safely 4.00 m above the river.

Answer 1

the kientic energy at 25 is 700*25=17500 joul

now , the required cord should make a strech of no more than 11 meter and store the above energy ,and then stop with 7 m strech.

d*f=energy
f=17500/11=1585N
so the cord should have a constant of no more than 11/1585=0.007m/N ( in order not to reach river)

at 4mete from river :
constant=d/weight=(32-25)/700= 0.01m/N

so the required constant should be 0.007m/N

Answer 2

Not sure of the definition of "force constant" but assuming it is the force per extension of the cord then:

elastic force from cord = force constant x (length of stretched cord - original length of cord)

force of gravity = weight of student

net force on student = force from cord - force of gravity

mass of student = weight / acceleration of gravity

acceleration of student = net force / mass

velocity of student = integral of acceleration with respect to time

position of student = integral of velocity with respect to time

You need to find the force constant that makes the velocity=0 when positon = 4m above the river.

I note that acceleration of gravity is not given in the question so if your answer is only slightly wrong maybe you have assumed the wrong value.

Alternatively you can do it the easy way by balancing the gravitational potential energy at the top ( = weight x height) with the elastic energy at the bottom (= integral of elastic force x distance). Then you don't need to know the acceleration of gravity.

Ensure you are not making any silly mistakes like forgetting the cord exerts no force until the student has fallen 25 m.

(Later)

I see PowerX has interpreted the question differently, i.e. has assumed the question means the final resting place of the student is 4m above the water. I assumed that was the position where the student initially came to rest at full extension of the cord, then bounced back up.

PowerX however made an error in the first part of his answer. He said that the cord at 11 m extension must store the kinetic energy that was available after a 25 m drop. In fact it must store the total energy available after a 36 metre drop. This will be equal to the initial gravitational potential energy, i.e. height x weight.

He has also defined the force constant as extension per unit force, rather than force per unit extension. You need to check if the answer in the book has units of m/N or N/m.

So doing it his way but using k = force per unit extension (N/m):

Potential energy at top = 36 x 700
Elastic energy at bottom = extension x k
= (36-25) x k = 11 x k
Equate the two energies
36 x 700 = 11 x k

Hence find k. (will be about 2300)

This is the minimum k to avoid getting wet.

To come to rest 4 m above the water, Power X had the right calculation

weight = elastic force = extension x k

700 = 7 x k

k=100 N/m

THis is less than the minimum safe value calculated above, so it is not possible to avoid the water AND finally come to rest precisely 4.0 m above the water. (Which probably means this interpretation of the question was the wrong one.)

Here the k is in N/m. If you need the answer in m/N it will be 1/k.

Answer 3

OK.
Do it based on the students potential energy and the potential energy the bungee cord will contain when the student comes to a stop (before she starts back up). These two numbers will be the same (from conservation of energy). The students potential energy through a drop of 36 - 4 = 32 meters is
Ep = Fh = 700*32 = 22400 J
In stopping her, the cord stretches 32 - 25 = 7 meters and, when fully stretched, it contains
Ep = kx²/2 J (where x is the length stretched) or
k = 2*(22400)/7² = 948.28 n/m.

HTH ☺

Doug