Physics Question?

Question

Physics Question?

You were so impressed with the problem about Super Dave that you decide that this would make a good stunt for the Institute of Technology (IT) day. To raise money for a University scholarship fund, you want to have the new IT dean bungee jump from a crane if contributions can be found for 10 scholarships. To add some interest, the jump will be made from 44 m above a 2.5 m deep pool of Jello. A 30-m long bungee cord would be attached to the dean's ankle. First you must convince the dean that your plan is safe for a person of his mass, 70 kg. The dean knows that as the bungee cord begins to stretch, it will exert a force which has the same properties as the force exerted by a spring. Your plan has the dean stepping off a platform and being in free fall for the 30 m before the cord begins to stretch. You must determine the elastic constant of the bungee cord so that it stretches only 12 m, which will just keep the dean's head out of the Jello.

Answer 1

The velocity of the dean after 30m of free fall is Sqrt(2*g*30) = Sqrt(588) taking g as 9.8m/s^2

Now, for the elastic, the motion is going to be simple harmonic, form the moment it starts stretching, till the dean reaches the bottom most point, and rebounds to the point where the stretching started.

The equation would be: x(t) = 30m + XSin(Sqrt(k/m)*t) where t = 0 when the elastic starts stretching and x is the distance from the top.

Now, X is the maximum stretching possible, that is 12m.

Differentiating the equation, we get v(t) = 12 cos(Sqrt(k/m)*t) * Sqrt(k/m)

since the velocity is Sqrt(588) when t = 0:
Sqrt(588) = 12 * 1 * Sqrt(k/70)
588/144 = k/70. => k = 285.83N/m.

Answer 2

Sure. The Dean 'falls' a total distance of 42 meters during which time gravity does 70*9.8*42 = 28,812 J of work. At the end of that distance, his velocity is 0 so all of the energy must be in the stretched bungee cord. The cord has stretched 12 meters and contains 28,812 J of energy and, since the energy contained in a stretched spring (or bungee cord☺) follows Hooks Law
E = (1/2)*k*x² = > k = 2*28,812/144 = 400.1666 J/m²

In practice..... Make it about 399 J/m². If he's stupid enough to trust his students (without double checking their work), he *deserves* to get a face full of Jello ☺


Doug

Answer 3

( i sure could use a scolar for a second master's degrees)

i like this question since you can visualize it as it happens.
so you have someone jumping, he accelerates till the moment the rope starts to stretch. and then he decelarates for a certain time and distance in order just to reach Jelo level.
let us pt it down to physics and equations.
as i see it you must slow down the pure man's speed he gathered while in a 30m free fall in 14meters distance.

Ok from 44m to 14m (30 meters free fall no force applied from the rope)

1/2mU^2 = mgh where H = 30m ==>>

U = sqrt (2*9.81*30) = 24.26 m/s


the tricky part of this question to be takes seriously into consideration is that the man will accelerate till his body weight equals the F of the rope that will slow him down.

the a from 14m to 0m is postive become zero and negative.nad at zero all the kinetic energy of the man and its potential energy should be transformed nto potential energy of the spring.
So,
1/2mu^2 + mgh = 1/2kx^2 + mgh1 where h=x = 12m. (in order to save the dean's head)

=> k=400.54






at least you got the main idea of the solution .
please tell me if it is right

Answer 4

I dont know. bullshit!