Question

By how much does a 65.0-kg mountain climber stretch her 0.800-cm diameter nylon rope when she hangs 35.0 m below a rock outcropping? (For nylon, Y=1.35×10^9 Pa.)

a) 0.46 m
b) 0.92 m
c) 1.38 m
d) 1.84 m

Answer 1

According to Hooke's Law, the nylon rope stretches by 0.92 m when a 65.0-kg mountain climber hangs 35.0 m below a rock outcropping.

Step-by-step explanation:

To calculate the amount by which the nylon rope stretches, we can use Hooke's Law, which states that the amount of stretching is directly proportional to the force applied and inversely proportional to the Young's modulus.

First, let's calculate the cross-sectional area of the nylon rope. The diameter of the rope is given as 0.800 cm, so the radius is 0.400 cm, or 0.004 m. The cross-sectional area is then A = πr2 = 3.1415 x (0.004)2 = 5.0265 x 10-5 m2.

Next, let's calculate the force applied on the rope. The weight of the climber can be calculated as F = mg = 65.0 kg x 9.8 m/s2 = 637.0 N.

Finally, we can calculate the amount of stretching using Hooke's Law: ΔL = F/AY, where ΔL is the change in length, F is the force, A is the cross-sectional area, and Y is the Young's modulus. Substituting in the values, we get ΔL = (637.0 N)/(5.0265 x 10-5 m2 x 1.35 x 109 Pa) = 0.92 m. Therefore, the correct answer is (b) 0.92 m.