Mountain Climber- how many cm will the rope stretch before she stops?

[huybinhs]

Homework Statement

A mountain climber is attached to a rope. She slips, and after she has fallen 4.5 m, the rope begins to help decelerate her. If the constant deceleration is 5.2g, how many cm will the rope stretch before she stops? (For most real ropes, the deceleration depends on the stretching of the rope.)

Homework Equations

x = x0 + v0t + 1/2 at2

v = v0 + at

2(x-x0) = (v+v0)t

v^2 = v^02 + 2a(x − x0)

g = 9.8 m/s^2

The Attempt at a Solution

mgh1 = mgh2
mass cancels on each side

(9.8m/s^2) * 5.2 = 50.96 m/s^2

(50.96 m/s^2) h = 9.8m/s^2 * 4.5 meter
h = 0.865 meters of rope stretch
= 86.5 cm

=> incorrect answer. Please help!

 

[huybinhs]

Anyone?

 

[huybinhs]

I still couldn't get the right answer! Help please!

 

[Saladsamurai]

Hi! First off, your problem statement could use some work:

constant deceleration is 5.2g

what does this mean? grams?

(9.8m/s^2) * 5.2 = 50.96 m/s^2

what are you calculating here?

Anyway, let's think about this in a different light: You know that there is a relationship between the Work done on the climber and the climber's change in kinetic energy right?

Can you give me an equation that represents this relationship?

 

[huybinhs]

Hi! First off, your problem statement could use some work:


what does this mean? grams?



what are you calculating here?

Anyway, let's think about this in a different light: You know that there is a relationship between the Work done on the climber and the climber's change in kinetic energy right?

Can you give me an equation that represents this relationship?

5g = 5 * 9.8 m/s^2.

The equation is included on my problem above ;)

 

[Saladsamurai]

Hi again! No. That is not an equation, that is a calculation. I am looking for the relationship between work and kinetic energy. It says that

W = something to do with KE

 

[huybinhs]

Hi again! No. That is not an equation, that is a calculation. I am looking for the relationship between work and kinetic energy. It says that

W = something to do with KE

Here is how I just did:

a) velocity when she falling:

v^2 = 0 + 2(9.8)*4 = 78.4 => v = 8.9 m/s

b) when she fall at the end of the rope then she stops, so velocity at that time is 0 which is v = 0 m/s.
Now we know that at the end of the rope her velocity is 8.9 m/s, then she stop and continue to falling down (the rope stretching) => v0 = 8.9 m/s. We need to find her velocity after the rope stretching:

0 = 8.9^2 + 2(-5.6)Delta x
=> Delta x = 7.07 m => 707 cm => wrong answer.

What am I doing wrong?