Question

Can you please help me solve this question with its all the parts?

(8%) Problem 9: The length of nylon rope from which a mountain climber is suspended has a force constant of 1.2 x 104N/m. > * 25% Part (a) What is the frequency, in Hz, at which he bounces, given his mass and the mass of his equipment is 84 kg? f=1 Grade Summary Deductions 0% Potential 100% sino 7 8 9 HOME JT ( E^^ 4 5 6 cotan atan cosh cos tan asino acos acotan sinh tanh cotanh() Degrees O Radians 2 3 Submissions Attempts remaining: 46 0% per attempt) detailed view 1 0% 2 0% 3 0% 4 0% + - 0 END Vol BACKSPACE DL. CLEAR Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback. A 25% Part (b) How much would this rope stretch, in centimeters, to break the climber's fall if he free-falls 1.2 m before the rope starts to stretch? A 25% Part (c) What is the frequency, in Hz, at which he bounces, given his mass and the mass of his equipment is 84 kg if the rope is twice as long? A 25% Part (d) How much would this rope stretch, in centimeters, to break the climber's fall if he free-falls 1.2 m before the rope runs out of slack if the rope was twice the length?

Answer 1

Given, forme constant of the rope, K=1.2*10*W/m m= 84 kg. til HZ 2. 2 gy 1.2x104 Hz Hz 84 frequeney = 1.g HZ Let us chise Point Zero at the equilibrium of the rope. So, before the elimber falls heis at a height Woove zero. h = 1.2m. . Potential Energy of the Climber U, = myho. E, Now, onu he falls, and the rope stops him lit on be the distanu upto which the rupe is Stritchet, So, no we have two energies, 1) Spring potential energh (drom Strutched reope) i Gravitational Potential energy (as climber is now at a distanu in below h =o). E = my (-2) + / km² sine, the energy is conserved here, so fifa > mgho / Kn? - myn 1 kn? - myn-mg ho knt amgn- 2 my ho = [as ho = 1.2m kn² amga 24 mg ang & Jumg 7 4. K. 24 my ak =x (mg & my 42.4k.my) 2 the chose the positive solution here. n

in= 0.435m. = 43.5 cm When rope length is doubled then, 0.6810" W/m. 4 K'= 1/2 - и! m • 1 345 HZ D) Following similar. Procedure we get, n'= 0.734m ~ 73.4em.