How much would the rope stretch to break the climber's fall if he free falls 2m before the rope runs out of slack? k=1.40x10^4N//m

mass of man= 90 kg
I forgot how to use conservation of energy for this. I tried it, but I did it wrong.

To determine the stretch in the rope using conservation of energy, you need to consider the potential energy and the work done on the rope. Here's how you can calculate it correctly:

1. Start by calculating the potential energy of the climber before the fall by using the formula:
Potential Energy = mass * gravitational acceleration * height
Potential Energy = 90 kg * 9.8 m/s^2 * 2 m

The potential energy before the fall is 1764 J.

2. Next, calculate the work done on the rope as the climber falls using the formula:
Work = Force * distance
In this case, the force is the spring constant (k) multiplied by the stretch in the rope (x).

The work done on the rope is equal to the potential energy of the climber, so we can equate these values:
Potential Energy = Work
1764 J = k * x

3. Now we can solve for the stretch (x) in the rope. Given that the spring constant (k) is 1.40x10^4 N/m:
x = Potential Energy / k
x = 1764 J / (1.40x10^4 N/m)

By plugging in the values, you will find that the stretch in the rope is approximately 0.126 m.

x = 1/2*g*t^2 = 2 = 4.9*t^2

Solve for t, the time then solve for the climber's speed v using

v = 9.8*t

1/2*m*v^2 = 1/2*k*xstretch^2; where xstretch is the amount that the rope will stretch