Question

The length of nylon rope from which a mountain climber is suspended has a force constant of 1.40×10^{4} 4 N/m . (a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg? (b) How much would this rope stretch to break the climber's fall if he free- falls 2.00 m before the rope runs out of slack? Hint: Use conservation of energy.

Answer 1

Answer:

Part a)

$f = 1.98 Hz$

Part b)

$x = 0.57 m$

Explanation:

Part a)

as we know that frequency of oscillation of the object connected to elastic rope is given as

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

now we have

$k = 1.40 \times 10^4 N/m$

m = 90 kg

now from above equation

$f = \frac{1}{2\pi}\sqrt{\frac{1.40 \times 10^4}{90}}$

$f = 1.98 Hz$

Part b)

If the climber falls freely till L = 2 m before rope start to stretch

then to find the maximum stretch we will have

$W_g + W_{rope} = 0$

$mg(L + x) - \frac{1}{2}kx^2$

$(90 \times 9.8)(2 + x) = 0.70\times 10^4 x^2$

by solving above equation we have

$x = 0.57 m$