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

1. Feb 17, 2010

### huybinhs

1. The problem statement, all variables and given/known data

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.)

2. Relevant 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

3. 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

2. Feb 17, 2010

### huybinhs

Anyone?

3. Feb 17, 2010

### huybinhs

4. Feb 17, 2010

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?

5. Feb 18, 2010

### huybinhs

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

The equation is included on my problem above ;)

6. Feb 18, 2010

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

7. Feb 18, 2010

### huybinhs

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?