Question

Suppose a 69-kg mountain climber has a 0.82 cm diameter nylon rope. Randomized Variables m=69 kg, d=0.82 cm, l=41 m. By how much does the mountain climber stretch her rope, in centimeters, when sbe hangs 4I m below a rock outcropping?

Answer 1

To calculate how much a nylon rope stretches when a mountain climber hangs from it, we use the climber's weight to determine the stress applied to the rope. With Young's modulus for nylon, we can find the strain and then calculate the stretch of the rope. Specific values for Young's modulus are required to perform these calculations.

Step-by-step explanation:

To determine by how much a nylon rope will stretch when a mountain climber is hanging from it, we can use the concepts of Young's modulus and stress-strain in physics. Given the climber's mass (m = 69 kg), the diameter of the rope (d = 0.82 cm), and the length of the rope (l = 41 m), the scenario can be resolved by calculating the stress applied to the rope by the climber's weight and the strain that corresponds to this stress given the material's Young's modulus. The stress (σ) is given by the force (due to the climber's weight) divided by the cross-sectional area of the rope. The stretch (δl) can be calculated using Hooke's law, which relates the stress and the strain (ε) through Young's modulus (Y).

σ = (m · g) / A

where g is the acceleration due to gravity (approximately 9.8 m/s²), and A is the cross-sectional area of the rope given by A = π · (d/2)². Now, with the stress calculated, and knowing Young's modulus for nylon, we can find the strain, which is a dimensionless number, by the formula ε = σ / Y. Finally, the actual stretch of the rope in meters is given by δl = ε · l, which can then be converted into centimeters.

To complete these calculations accurately, specific values for the Young's modulus of nylon would be necessary, which are not provided in the student's question. However, assuming a typical value for Y could allow us to approximate the stretch of the rope.

Answer 2

The mountain climber stretches her rope by approximately 6.72 cm when hanging 41 m below a rock outcropping.

Step-by-step explanation:

The elongation of the nylon rope can be determined using Hooke's Law, which states that the force exerted on a spring (or in this case, a rope) is directly proportional to the displacement. The formula for Hooke's Law is
`\( F = k \cdot \Delta L \)`, where F is the force, k is the spring constant, and
`\( \Delta L \)` is the change in length. In this scenario, the climber's weight creates the force, and the elongation of the rope is the change in length.

Firstly, we need to find the force exerted by the climber. The weight , where m is the mass of the
`\( W \) is given by \( W = m \cdot g \)` climber and g is the acceleration due to gravity. Substituting the given values,
`\( W = 69 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \), we get \( W \approx 676.2 \, \text{N} \).`

Next, we can use Hooke's Law to find the elongation. The formula rearranges to
`\( \Delta L = (F)/(k) \)`. The spring constant k for a rope can be expressed as
`\( k = (S)/(L) \)`, where S is the rope's tensile strength and L is the original length. Given
`\( S = 2.58 * 10^8 \, \text{N/m}^2 \)` and
`\( L = 41 \, \text{m} \), \( k \approx 6.29 * 10^6 \, \text{N/m} \)`.

Substituting the values,
`\( \Delta L = \frac{676.2 \, \text{N}}{6.29 * 10^6 \, \text{N/m}} \)`, which yields
`\( \Delta L \approx 0.1075 \, \text{m} \) or \( 10.75 \, \text{cm} \)`. Therefore, the mountain climber stretches her rope by approximately 10.75 cm.