A 74.2-kg climber is scaling the vertical wall of a mountain. His safety rope is made of a material that, when stretched, behaves like a spring with a spring constant of 1.62 x 103 N/m. He accidentally slips and falls freely for 0.959 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

The potential energy going into the rope is .959+x meters (x is the distance the rope stretches).

the stored energy in the rope is 1/2 k x^2

set them equal, solve for x.

To solve this problem, we need to calculate the amount of rope stretching required to bring the climber to rest.

We can use the concept of energy to solve this problem. When the climber falls, the potential energy of the climber is converted into the potential energy stored in the stretched rope.

The potential energy stored in a spring is given by the formula:

PE = (1/2) * k * x^2

Where:
PE is the potential energy stored in the spring (rope)
k is the spring constant
x is the amount of rope stretching

Given:
The mass of the climber (m) = 74.2 kg
The spring constant (k) = 1.62 x 10^3 N/m
The distance fallen (d) = 0.959 m

To bring the climber to rest, the potential energy stored in the rope should be equal to the gravitational potential energy of the fallen distance.

Gravitational potential energy (PEg) can be given by:

PEg = m * g * h

Where:
m is the mass of the climber
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the fallen distance

Now we can equate the potential energy stored in the spring to the gravitational potential energy:

(1/2) * k * x^2 = m * g * h

Plugging in the given values:

(1/2) * (1.62 x 10^3 N/m) * x^2 = (74.2 kg) * (9.8 m/s^2) * (0.959 m)

Simplifying:

(1.62 x 10^3 N/m) * x^2 = (74.2 kg) * (9.8 m/s^2) * (0.959 m) * 2

x^2 = ((74.2 kg) * (9.8 m/s^2) * (0.959 m) * 2) / (1.62 x 10^3 N/m)

x^2 ≈ 3.564 m^2

Taking the square root of both sides:

x ≈ √3.564 m

x ≈ 1.885 m

Therefore, the rope will stretch approximately 1.885 meters when it breaks the climber's fall and momentarily brings him to rest.

To find how much the rope is stretched when it breaks the climber's fall, we need to determine the amount of potential energy stored in the stretched rope when it momentarily brings him to rest. We can calculate this using the formula for the potential energy stored in a spring:

Potential Energy = (1/2) * k * x^2

Where:
k = spring constant
x = displacement from the equilibrium position

Given:
Mass of the climber (m) = 74.2 kg
Spring constant (k) = 1.62 x 10^3 N/m
Displacement (x) = 0.959 m

Now, let's calculate the potential energy stored in the stretched rope:

Potential Energy = (1/2) * k * x^2
= (1/2) * (1.62 x 10^3 N/m) * (0.959 m)^2

Calculating this expression:

Potential Energy = 717.122 J

Therefore, when the rope breaks the climber's fall and momentarily brings him to rest, the rope is stretched and stores approximately 717.122 J of potential energy.