Question

A bungee rope is stretched to twice its normal length by a person of mass 75 kg hanging at rest from the free end (which is 70m below the platform).

For a person of mass m kg, calculate the depth to which a person would fall if attached
to a rope of the type described above, with length L in metres. (Neglect the jumper's height.)

This is a PHYSICS assignment for Mathematics C ... I can't do physics :3

Answer 1

we can treat the bunjee rope like a spring

Answer 2

That's understandable. So the force net = mg - kx where k is the coefficient of elasticity for the rope and x is the extension of the cord, in metres. But where to from there? 'k' and 'x' are undetermined. (Sorry for my late response)

Answer 3

looks good, -kx comes from hooks law

Answer 4

the force of tension

Answer 5

So what would be the recommended steps with this formula and the provided information in order to determine the formula for the length of rope in respect to mass?

Thank You

Answer 6

Let me try to help you out. :) So we would have to use F net = mg - kx twice. First time, to find the value for k and then the second time, to solve for x
But in both cases... what would F net be equal to, when the person is at the lowest point?

Answer 7

We have to think about what F net is at the lowest point in the motion
By any chance, have you done Free Body Diagrams? o.O

Answer 8

Sorry. I have dinner. gtg. thanks

Answer 9

Enjoy your dinner

Answer 10

You can equate the decrease in gravitational potential energy to the increase in the elastic stored PE of the rope. As in \( \bf 1/2 k x^2 = m g (x+L) \), where x is the extension of the rope beyond it's unstretched length and k,L are constants that can be determined from the given data.

Answer 11

Hey guys. Thanks for the help. I think I may have cracked something. But I need to know how to integrate 'v(dv/dx)'.

Answer 12

Perl: Are not 'L' and 'k' the same thing? 'k' is the spring's constant. I thought this meant it was the length of the rope?

Answer 13

Micracrown: Um ... I'll just smile and nod my head :P I have no idea what 'Free Body Diagrams' are. Though it sounds like a complex name given to a simple thing :P I have no idea about physics (even the darn words confuse me.) I feel like the lowest point in the motion would be where v = 0? If so, this would be when the participant is standing on the platform and when they are hanging at rest.

Answer 14

Perl: I don't know how to apply your formula. This is what I did.

(kx^2)/2 = mg(x + L)

F = ma, ma = kx. 75 x 9.8 = 35k. (75 x 9.8)/35 = k = 21.

x + L = 70 (For a Dry Jump). Therefore, x = 70 - L.

(21 x (L/2)^2)/2 = 75 x 9.8(L/2 + L)
(21L^2)/4 = 735(3L/2)
5.25L^2 - 1,102.5L = 0
Quadratic equation. L = 106.16 or 0. These are both unreasonable numbers.

What have I done wrong?

Answer 15

@Michele_Laino can you take a look at this

Answer 16

If I call with L the length of the rope at the rest, then I can write:
\[\large 75 \times 9.81 = K\left( {70 - L} \right)\]
that is the condition of equilibrium, and K is the rope constant.
Now from the above equation I get:
\[\large K = \frac{{75 \times 9.81}}{{70 - L}} = \frac{{737.75}}{{70 - L}}\]

Answer 17

now the relationship between the depth x and mass m is :
\[\large mg = K\left( {x - L} \right)\]
where g is the gravity, namely g=9.81 m/sec^2

Answer 18

So, substituting K into that last formula, we get:
\[\large x = \frac{{mg}}{K} + L = \frac{{m \times g}}{{75 \times 9.81}}\left( {70 - L} \right) + L = \frac{m}{{75}}\left( {70 - L} \right) + L\]

Answer 19

Thanks for calling in reinforcements Perl :P :)
Michele: How do I use that result? 'x' equals '(m/75)(70-L) + L' which expanded is '(12/13)m - (L/75)m + L'. But what can we do with this if we're trying to solve 'L' and not 'x'? Just looking at it I don't think we are able to isolate L by itself?

Answer 20

x is the depth, due to the mass m, whereas L is the length of the rope at rest, namely with no mass applied on it

Answer 21

x is a variable, whereas L is a parameter

Answer 22

But we're not trying to find the depth due to m. The depth will always be 70m for a dry jump.

Answer 23

and 72m for a wet jump.

Answer 24

it is 70 meters when a mass of 75 Kg is applied to your rope

Answer 25

whereas when the applied mass has a generic value, namely m, the depth is not 70 meters, but a generic value x

Answer 26

Yeah. Sorry mate. I understand the confusion as it is how I previously felt. The way the question is phrased (which I have taken from the assignment) is misleading. The depth to which they fall will always be 70m for a dry jump and 72m for a wet jump. The assignment wants us to find the formula that determines the length of rope required to ensure this despite the person's mass. Sorry about that. But I know that the length of rope with respect to mass is the desired formula ... and that the assignment is just worded STUPIDLY. Thanks though :)

Answer 27

Thanks! :)

Answer 28

ok!