By how much does a 64.3 kg mountain climber stretch her 0.910 cm diameter nylon rope when she hangs 32.2 m below a rock outcropping? (For nylon, Y = 1.35 â 10^9 Pa. Enter your answer in centimeters.)

Question

Adrien Taylor

Physics

9 July, 18:23

By how much does a 64.3 kg mountain climber stretch her 0.910 cm diameter nylon rope when she hangs 32.2 m below a rock outcropping? (For nylon, Y = 1.35 â 10^9 Pa. Enter your answer in centimeters.)

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Answers (1)

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Erick Larson · Answer 1

23 cm

Explanation:

m = 64.3 kg, diameter = 0.910 cm

radius, r = 0.455 cm = 4.55 x 10^-3 m

A = 3.14 x (4.55 x 10^-3) ^2 = 6.5 x 10^-5 m^2

L = 32.2 m

Y = 1.35 x 10^9 Pa

Let the stretched length is l.

Use the formula for Young's modulus

Y = mg L / A l

l = m g L / A Y

l = (64.3 x 9.8 x 32.2) / (6.5 x 10^-5 x 1.35 x 10^9)

l = 0.23 m

l = 23 cm