Rock climber elastic rope problem

[hewhoeatspie]

SOLVED

Homework Statement

Two rock climbers, Bill and Karen, use safety ropes of similar length. Karen's rope is more elastic, called a dynamic rope by climbers. Bill has a static rope, not recommended for safety purposes in pro climbing. Karn falls freely about 2.0m and then the rope stops her over a distance of 1.0m. (a) Estimate, assuming that the force is constant, how large a force she will feel from the rope. (Express the result in multiples of her weight.) (b) In a similar fall, Bill's rope stretches by 30cm only. How many times his weight will the his weight will the rope pull on him? Which climber is more likely to be hurt?

Homework Equations

(Not entirely sure)
d=g(0.5)t^2
a=v/t

The Attempt at a Solution

d=g(0.5)t^2
1.0=9.8(0.5)t^2 = t=0.45
(I used a simulation to get that something that falls for 2m will be falling at a velocity of 6.2 m/s...I'm not the best at physics.)
v=6.2m/s
a=(-6.2m/s)/0.45 = a=-13m/s^2
F=Frope-Fg
ma=Frope-Fg
Frope=ma-Fg
Frope=m(-13m/s^2)-Fg
?

That that's where I got stumped. I'm sorry if I didn't provide enough information, or if it was jumbled, I'm new here.

I haven't even tried the second half...

EDIT: Nevermind, I got some help and solved it.

 

[Nate810]

a) Karen will feel a force of 13 times her weight.b) Bill will feel a force of 8.7 times his weight. Since Karen's rope is more elastic, she is less likely to be hurt.

 

[baba89]

I can provide a more thorough and accurate solution to the rock climber elastic rope problem. Firstly, we can use the equation F=ma to determine the force experienced by Karen when she falls. We know that her mass is equal to her weight divided by the acceleration due to gravity (m=mg), so we can rewrite the equation as F=mg*a. Since we are given the distance and velocity of her fall, we can use the equation d=vt-0.5at^2 to solve for the acceleration (a) which is equal to -13m/s^2. Plugging this into our original equation, we get F=mg*(-13m/s^2) which simplifies to F=-13mg. This means that Karen is experiencing a force that is 13 times her weight, or 13g.

For Bill, we are given that his rope stretches by 30cm, or 0.3m, when he falls. We can use the same equation d=vt-0.5at^2 to solve for the acceleration (a) which is equal to -0.3m/s^2. Plugging this into our original equation, we get F=mg*(-0.3m/s^2) which simplifies to F=-0.3mg. This means that Bill is experiencing a force that is 0.3 times his weight, or 0.3g. This is significantly less than the force experienced by Karen, making her more likely to be hurt in a fall.

In conclusion, it is important for climbers to use dynamic ropes, which are more elastic and can absorb more energy, thus reducing the force experienced during a fall. This is crucial for safety purposes in professional climbing.