Falling Rock Climber (Work and Energy)

[kubaanglin]

Homework Statement

Stretchable ropes are used to safely arrest the fall of rock climbers. Suppose one end of a rope with unstretched length $l$ is anchored to a cliff and a climber of mass m is attached to the other end. Ehrm the climber is a height $l$ above the anchor point, he slips and falls under the influence of gravity for a distance $2l$, after which the rope becomes taut and stretches a distance $x$ as it stops the climber. Assume a stretchy rope behaves as a spring with spring constant $k$.
(a) Applying the work-energy principle, show that
$$x=\frac{mg}{k}\left[1+\sqrt{1+\frac{4kl}{mg}}\right]$$

Homework Equations

$$U_{GPE}=mgh$$
$$K=\frac{1}{2}mv^2$$
$$U_{spring}=\frac{1}{2}kx^2$$
$$F_{spring}=|kx|$$
$$W=ΔK$$

The Attempt at a Solution

Before the climber falls

$U_{GPE}=mg(2l+x)$ and $U_{spring}=0$

After the climber has reached maximum displacement

$U_{GPE}=0$ and $U_{spring}=\frac{1}{2}kx^2$

Therefore $mg(2l+x)=\frac{1}{2}kx^2$

I feel that this is incorrect because I am not using the work-energy principle, but what is the relevance if the kinetic energies at the beginning and end of the fall are zero? Should I split this problem into two parts where $\frac{1}{2}mv^2=mg(2l)$ at the point when the rope begins to stretch and then work from there?

 

[gneill]

You've done alright. You've used the work-energy principle to convert the GPE into the work done by gravity resulting in KE of the climber. You didn't, for example, use kinematic equations and the acceleration due to gravity to find the speed, then determine the KE from that.

Just solve your equation for x.