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 calculate how much the rope stretches to break the climber's fall, we can use the principle of conservation of energy. The potential energy lost by the climber as he falls will be converted into the elastic potential energy stored in the stretched rope.
Let's break down the problem into steps:
Step 1: Calculate the potential energy lost by the climber.
Given that the climber falls 2m, we can calculate the potential energy lost using the formula:
Potential Energy (PE) = mass (m) x gravity (g) x height (h)
PE = 90 kg x 9.8 m/s^2 x 2 m
Step 2: Calculate the elastic potential energy stored in the rope.
The elastic potential energy (EPE) stored in the stretched rope can be calculated using the formula:
EPE = (1/2)kx^2
Where:
- EPE is the elastic potential energy stored in the rope
- k is the spring constant of the rope (given as 1.40x10^4 N/m)
- x is the stretch or extension of the rope
Step 3: Equate the potential energy lost to the elastic potential energy stored in the rope.
Since energy is conserved, we can equate these two values:
PE = EPE
90 kg x 9.8 m/s^2 x 2 m = (1/2)(1.40x10^4 N/m) x x^2
Simplifying the equation gives us:
1764 N/m x 2 m = 0.7x^2
3528 N = 0.7x^2
Step 4: Solve for the stretch (x) of the rope.
Rearranging the equation, we get:
x^2 = 3528 N / 0.7
x^2 = 5040 N
x = √(5040 N)
Therefore, the rope would stretch approximately √(5040 N) meters to break the climber's fall.
Note: Make sure to calculate and verify the numerical solution based on the given values and units provided in the question.