Question

Suppose a 61-kg mountain climber has a 0.74 cm diameter nylon rope. Randomized Variables m=61 kg d=0.74 cm l=41 m ​ △ By how much does the mountain climber stretch her rope, in centimeters, when she hangs 41 m b of the rope is 5×10 9 N/m ^2 .

Answer 1

We calculate the stretch of a nylon rope under the weight of a mountain climber by using Hooke's law and Young's modulus, accounting for the force due to the climber's weight, the cross-sectional area of the rope, and the original length of the rope.

Step-by-step explanation:

To calculate how much a nylon rope stretches when a mountain climber hangs from it, we use Hooke's law and the definition of Young's modulus. We are given that a 61-kg mountain climber is hanging from a rope that has a diameter of 0.74 cm and a Young's modulus (Y) of 5×109 N/m2. The elongation of the rope (Δl) can be calculated by rearranging the formula for Young's modulus, Y = (F/A) / (Δl/l), where F is the force applied (weight of the climber), A is the cross-sectional area of the rope, l is the original length of the rope, and Δl is the change in length (stretch).

First, we calculate the force F by multiplying the mass of the climber by the acceleration due to gravity (g = 9.8 m/s2). Next, we find the cross-sectional area of the rope by using the formula for the area of a circle (A = πr2, with radius r = diameter/2). Finally, we rearrange the Young's modulus formula to solve for Δl, yielding Δl = (F×l) / (A×Y).

Using these calculations, we can find the stretch of the nylon rope to answer the student's question with precision and provide them with a detailed understanding of the physics concepts involved.