Physics Question, i just can't get it.?

Question

A 100kg bungee jumper jumps from a bridge 40m above a river. The bungee cord has an unstretched lenght of 18m and has a spring constant k = 200 N/m
a) How far above the river is he when the cord brings him to a stop?
b) What maximum force does the cord exert on him?

Full working would be appreciated (or the gist of it) Thanks in advance

Answer 1

Its a conservation of energy question. A classic one for introductory mechanics. This question really helped me learn about the problem solving process when I was a freshman physics student although your version is a bit less complicated. So I will not give you the answer. But I will tell you that the final equation should come out to be quadratic and you can solve form there.

You will remember this problem for a long time if you get it on your own.

Answer 2

First, note that James H's analysis is incorrect. He's assuming that the max force exerted by the cord is equal to the jumper's weight. This is not the case. (If it were, the jumper would not "bounce back up" after reaching the maximum stretch.)

a) How far above the river is he when the cord brings him to a stop?

I would look at this in terms of conservation of energy. If you ignore losses due to heat and friction, then KE+PE should be the same at every point along the journey.

We'll use the fact that the PE in the bungee cord is:
PE_b = kx²/2

where "x" is the amount of stretch. But the cord doesn't start stretching until it reaches 18m. So for example, if its length is 19m, its stretch (x) is only 1m. In general, x = (L-18m), where L is the cord's length. So:

PE_b = k(L-18m)²/2

Now, at the top of the bridge, all the PE is gravitational potential energy, and there is no KE (because no motion). So:

E_initial = PE_initial + KE_initial
= mg(h_initial) + 0
= mg(40meters)

At the maximum stretch, again there is no KE (since he's momentarily stopped). The PE is a combination of gravitational PE plus the PE in the stretched cord:

E_final = PE_final + KE_final
= mg(h_final) + k(L-18meters)²/2 + 0

But notice that:
h_final + L = 40meters (draw a diagram if you don't see why)

or:
L = 40meters - h_final

so:
E_final = mg(h_final) + k((40meters - h_final)-18meters)²/2
= mg(h_final) + k(22meters - h_final)²/2

And, since E_initial = E_final:
mg(40meters) = mg(h_final) + k(22meters - h_final)²/2

This is a quadratic equation in h_final, so you can use the quadratic formula to find h_final. This will give you two solutions, one above 22 meters (the point where the cord starts to stretch) and the other below 22 meters. You can discard the higher value.

b) What maximum force does the cord exert on him?

By the definition of the spring constant k, this is just:
F_cord = k(L-18meters)

Previously we pointed out that L = 40meters-h_final, so:
F_cord = k(40meters - h_final - 18meters)
= k(22 meters - h_final)

Use the value of h_final calculated in part "a".

Answer 3

You must use the force/displacement equation, F = k*x, where k is the spring constant and x is the spring displacment (the amount the spring is stretched from it's equilibrium position). The jumper's MASS is 100 kg, while the FORCE the jumper exerts on the cord is governed by Newton's 2nd Law F= m*a. That is, F = (100 kg)*(9.81 m/s^2) = 981 N. To find the amount the cord stretches, use F = k*x, 981 N = (200 N/m)*x, x = 4.905 m. So, the spring stretches 4.905 meters, use that to figure how close the jumper comes to the river. We have already found the maximum force on the jumper by using newtons 2nd law.