The length of nylon rope from which a mountain climber is suspended has an effective force constant of 1.40 × 104 N/m.

Question

THIS IS THE FULL QUESTION.

The length of nylon rope from which a mountain climber is suspended has an effective force constant of 1.40 × 104 N/m. (a)What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg? (b) How much would this rope stretch to break the climber’s fall if he free-falls 2.00 m before the rope runs out of slack? (Hint: Use conservation of energy.) (c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used.

Answer 1

Hi Muhammad!

 

(a) We are meant to treat this situation as a mass (the climber and equipment) on the end of a spring (the rope).  For that, there is a formula for the frequency of oscillation (f).

 

f  = (1/2π)√(k/m)

 

where

 

f = frequency (Hz)

k = force (or spring) constant (N/m)

m = mass on the spring (kg)

 

In this case, all of these are given directly, and you can plug them right in.  Just let me know if you would like to check an answer!

 

(b)  Because only gravity and the spring force (from the rope) are the only forces doing work, mechanical energy (the sum of kinetic and ALL potential energies) is conserved through this entire process.

 

If the climber starts in free fall, then -- right when he is "dropped" -- all of his mechanical energy is gravitational potential (because he has not yet started moving -- no kinetic energy -- and the rope is still relaxed -- no spring potential energy).  So:

 

ME_o = KE_o + GPE_o + SPE_o = 0 + 0 + mgh_o

 

We can reference h = 0 to any point we want.  I think it is most convenient, given the information in the problem and what we are trying to solve for, to choose h = 0 at the point just where the rope starts to stretch (when its slack runs out).  This would mean h_o = 2 m.  So:

 

ME_o = mg(2 m)

 

Once the climber has fallen as far as the rope will let him, he is once again brought to rest, so his kinetic energy is 0 again.  At this point, he has two parts to his mechanical energy:  a GPE part and a SPE part.

 

ME_f = KE_f + GPE_f + SPE_f = 0 + mg(-x) + (1/2)kx²

 

where x is the distance the rope has stretched.  This means there is now spring potential energy (from the stretched rope) and gravitational potential energy (since the climber is a distance "x" below h = 0; this makes the GPE_f negative, since the climber is now BELOW h = 0, but that is allowed)

 

Now, because mechanical energy is conserved:

 

ME_o = ME_f

mg(2 m) = mg(-x) + (1/2)kx²

 

We are trying to solve for "x," so this can be rearranged into:

 

(1/2)kx² - mgx - mg(2 m) = 0

 

which is a quadratic equation in x.  You can then use the quadratic formula to solve for x.  You will likely get two solutions, one positive and one negative.  Since we have implicitly taken the x value above to be positive, you want to choose the positive solution.

 

I hope this helps!  Let me know if you have any questions at all about this, or would like to check an answer.

(c) Notice that at no time during this process did we care what he overall length of the rope was (as long as the climber and equipment still only free-fall 2 m before the rope starts to stretch, as seems to be indicated). That may give you a clue about Part (c), but let me know if you want to talk about this further.