Question

a mountain climber of mass 60 kg slips and falls a distance of 4. m at which time he reaches the end of his elastic safety rope. The rope then stretches and additional 2m before the climber comes to rest. What is the spring constant of the rope, assuming it obeys hooke's law

Answer 1

What's hooke's law for starters?

F=kx

Where F is the force of the spring, x is the distance, and k is the spring constant. So to solve for k, we just divide both sides by x.

F/x=k

Now we have a formula to solve for the spring force constant. Do we know the distance, x? Do we know the force?

Let's look, he's falling, and what do we know about that? F=ma. So lets plug that into our original equation.

(ma)/x=k

Now we're getting somewhere. Do we know the acceleration? Do we know the mass? Do we know the distance?

Answer 2

Wait, I'm not sure if this is completely right. I think this involves potential energy. It's been a while since I've done one of these let me think for a sec.

Answer 3

Here's the real way to solve this problem.

So at the top of the fall he has a lot of potential energy and at the bottom of his fall all of that potential energy is converted into kinetic energy.

So what's potential energy? U=mgh. What's kinetic energy? 1/2kx^2. These are just the integrals of the force with respect to x of F=mg and F=kx, nothing too crazy here.

So we know the initial potential energy is equal to the final kinetic energy, so we set them equal.

\[mgh=1/2kx^2\]and from here we can solve for k, the spring force constant. You know the mass, acceleration due to gravity, and the total height fallen by the climber from the most potential energy to the very bottom when he comes to rest before springing back up. So in this problem, h=x.

Answer 4

so k would equal 196?