A mountain climber of mass 60 kg slips and falls a distance of 4 m, at which tmie he reaches the end of his elastic safety rope. The rope then stretches an additional 2 m before the climber comes to rest. What is the spring constant of the rope, assuming it obeys Hooke's law?(adsbygoogle = window.adsbygoogle || []).push({});

a) 1.76*10^3

b) 3.52*10^3

c) 5.28*10^3

d) 7.04*10^3

I would think that F = kx = mg, where k(2) = (60)9.8 and k = 294 N/m, but that's not one of the choices. Alternatively, I thought that kx = (1/2)mv(f)^2 where v(f)^2 = 2gx and v(f) = 8.85 m/s, but that gives k = 1.17*10^3, which isn't a choice either.

Help please! It would be great if I could get help on this by tomorrow morning.

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# Ideal Spring Problem

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