the length of a nylon rope from which a mountain climber is suspended has a forc
ID: 1508070 • Letter: T
Question
the length of a nylon rope from which a mountain climber is suspended has a force constant of 1.35 x 10^4 N/m.
A) What is the frequency, in hertz, at which he bounces given his mass of his equipment is 94 kg?
B) How far, in CENTIMETERS, would the rope stretch to break the climber's fall if he free falls 1.2 m before the rope runs out of the slack?
C) Assume that tying the two ropes of the same length toffether would have the same effect on the spring constant as two springs connected in a series. What would the frequency of oscillationbe, in hertz, in the rope was twice as long?
D) If the climber falls the same distance before running out of slack, how far would the rope stretch, in CENTIMETERS, if it was twice as long?
Please do not give answers in meters. I need centimeters. Thanks!
Explanation / Answer
k = 1.35 * 10^4 N/m
m = 94 Kg
(a)
f = (1/2)*(k/m) = (1/2)*(1.35*10^4)/94)
f = 1.91 Hz
(b) PE gets converted into spring energy U. If the rope stretches "x," then
mg(1.2 + x) = 1/2*kx²
94*9.8 * ( 1.2 + x ) = 1/2 * 1.35 * 10^4 * x²
x = 0.478 m
x = 47.8 cm
(c)
Now, we have two springs in parallel, and
keq = k1 + k2 = 2*k = 2*1.35 * 10^4 N/m
f = (1/2)*(k/m) = (1/2)*(2*1.35*10^4)/94)
f = 2.70 Hz
(d)
mg(1.2 + x) = 1/2*kx²
94*9.8 * ( 1.2 + x ) = 1/2 * 2*1.35 * 10^4 * x²
x = 0.322 m
x = 32.2 cm
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.