Weekly physics help thread

[Nate]

Hi. Me here. Again. Because apparently I suck at physics.

*sighs*

A 60-kg bungee jumper jumps from a bridge. She is tied to a bungee cord that is 12m long when unstretched, and falls a total of 31 m.

The first thing we had to do was calculate the spring constant, k. No problem. I actually did that part easily, it's 100 N/m.

The SECOND part is what I don't understand. It says, "Calculate the maximum acceleration experienced by the jumper."

The back of the book says 22 m/s^2. But...how is she falling faster than gravity? In fact, to calculate the spring constant k, you had to state that her accleration was g. So I don't understand how in the world they got that.

[Cap'n Nick]

Well, if they don't mean gravity, there's only one other acceleration occuring here, and that's the deceleration of the jumper by the bungee cord. Could that be it?

[SP1]

Remember she doesn't stop at the bottom. If the spring force from the bungee (F = kx, where x = distance of the stretch) is to be strong enough to slow her to a stop AND fling her back up into the air, it will have to be larger than g, since if it was = g then she would just fall at constant velocity.

In order to reverse course, the net force up (spring minus gravity or kx - mg) must be positive, so kx >mg. Since it is likely to fling her up to about where she started (you've see these, right?), kx - mg, at the bottom, is going to be something like mg. That is, kx is about 2mg.

Do a force balance equation for this: Fr = Fg + Fb
where Fr = resultant for (Fr = ma)
Fg = gravity (Fg = -mg)
Fb = kx (watch the sign of x here)

Remember you get 12m of freefall to start the problem. To make the calculation easier, you could let the problem start at the 12m point, except that v(0) is not zero, but negative (down). x(0) = 0 and increases to 19m before velocity is zero and then reverses.

Try this out, if I haven't confused you too much. PM me if you need clarification.

[Warrior4Christ]

Just curious Kae, what are you studying?

[SP1]

OK, simpler this time:

At the bottom, velocity is zero. So acceleration is strickly based on two forces: Bungee and gravity.

[SPOILER]Bungee force (up) is Kx or 100 N/m times the 19m the cord is stretched = 1900N. On a 60 kg body (F = ma), this is a 31.67 m/ss acceleration up. However, gravity (down) subtracts from this at 9.8 m/ss for a net of 21.87 m/ss (up) or about 22 m/ss.

Did I miss some of the problem? How did you calculate that the spring constant was 100 N/m? If do this as an energy balance (the problem consists of the potential energy of the person, potential energy of the spring, and the kinetic energy of the person). At the top and bottom, KE = 0. At the top PEs = 0 and at the bottom PEp = 0. Conservation of energy says that KE+PEs+PEp = constant. So, PEp = mgh or 60 kg * 9.8 m/ss * 31m or 18,228 Nm. That means (at the bottom) that PEs = 18,228 Nm. Work is the integral of Force times distance, so PEs = integral(kx)dx or PEs = 1/2k(x*x). This gives me k = 101, which I suppose is close enough. [/SPOILER]

Did you do it this way?

[Nate]

Heh, condensed version since I've got to get to history.

First, the cord won't have any potential energy until after it hits 12 m. So I started the problem when the girl hits 12 m.

At that point, the only energy is going to be the kinetic energy of the girl, since the cord won't have any potential energy until after the 12 m mark. So the total energy at 12 m is going to be just 1/2 m(v^2).

Find velocity by the formula (v[final]^2) = (v[initial]^2) + 2a(y[final]-y[inital])

Initial velocity is zero, acceleration is 9.8, final x is 12 and initial is 0. Plug it in and you get 15.3 m/s as her velocity at 12 m.

Now, when she hits 31 m, she will have no kinetic energy, but she will have potential energy (mgy) and so will the bungee cord (1/2 k[x^2]).

Because of the law of conservation of energy, the energy at all points is equal, so you have:

1/2 m(v^2) = 1/2 k(x^2) + mgy

(Noting that the y for the potential energy in the right side is going to be a negative 19 m because it is below the reference point, which is 12 m)

Anyway, after plugging all the values in, I got around 100 N/m for the spring constant of the cord, which the book says is correct, so I guess I did it right. ^^