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Given that:

  • bungee jumper weighs 700N
  • jumps off from a height of 36m
  • needs to stop safely at 32m (4m above ground)
  • unstretched length of bungee cord is 25m

Whats the force required to stop the jumper (4m above ground)

First what equation do I use?

$F = ma$? But even if $a = 0$ $v$ may not equals 0 (still moving)

$W = F \Delta x$? Can I say if $\Delta x = 0$ object is not moving? Even then, I don't know the work ...

I tried doing:

$-32 = \frac{1}{2} (-9.8) t^2$

$t = 2.556s$

Then I'm stuck ... I know $t$ but I cant seem to use any other equations... $v_f, v_i =0 $

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Think about potential energy. – Colin K Oct 12 '12 at 12:57
up vote 2 down vote accepted

As others here have pointed out, the force of the bungee cord would vary, increasing as it is stretched.

So your question is not well posed. If this is an actual homework problem I would guess that you misread it and you are actually being asked to find the force constant of the bungee cord (assuming, as I will below, that it obeys Hooke's law). Or perhaps you want the maximum force on the jumper due to the bungee cord (which would be when it is stretched the most). Here is how you would get those...

Ignoring air drag, the only forces on the jumper are due to the spring (bungee cord) and gravity, both of which are conservative forces, so you have:

$$ \frac{1}{2} m v_{f}^{2} + \frac{1}{2} k x_{f}^{2} + m g y_{f} = \frac{1}{2} m v_{i}^{2} + \frac{1}{2} k x_{i}^{2} + m g y_{i} $$

Going from the start of the fall of the jumper to when the stretch of the spring is maximum, the initial speed and final speed of the jumper are zero, and the initial x is zero (spring is unstretched at the start), while you can choose the final y to be zero, leaving:

$$ \frac{1}{2} k x_{f}^{2} = m g y_{i} $$

Which can be simply interpreted as saying that the initial gravitational potential energy of the jumper-earth system ends up stored in the spring as elastic potential energy. From this equation you can get $ k $. You can then go on to find the force due to the spring on the jumper when it is stretched by any amount using $ F_{x} = - k x $, and in particular, the maximum force due to the spring.

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Well, I guess start by forgetting that the bungee is a spring, and would apply a non-constant force. But we'll ignore that first and imagine the bungee applies a constant force.

You jump off the bridge at 36m, plunge 25m to 9m from the ground, which leaves you 5m to come to a stop.

So, we can use the equations of constant linear motion to compute how fast you're going the moment the bungee tightens up: $v^2 = v_0^2 + 2a(r-r_0)$

So, in our example you'll be heading downwards at $\sqrt(0+2*9.8\frac{m}{s^2}(25m)) = 22.14\frac{m}{s}$

Then, assuming the bungee applies a constant force, we again use the initial equation to figure out the rate of deacceleration.

$0 = (22.14 \frac{m}{s})^2 + 2a(5m) => a= \frac{(22.14 \frac{m}{s})^2}{2*5m} = 49 \frac{m}{s^2}$

Which, not so surprisingly, works out to be the same as $g(25m/5m)$, or $g$ times the falling distance divided by the stopping distance. Now that you know your deacceleration, multiply that by your mass and you've got your force.

However, bungees actually don't apply a constant force, they apply a fairly linear force relative to their displacement for most of their stretchy range. You'll have to use Hooke's Law, the formula for the spring constant $F=-kx$ to more accurately model the system.

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