A mountain climber of mass 60.0 kg slips and falls a distance of 4.00 m, at which time he reaches the end of his elastic safety rope. The rope then stretches an additional 2.00 m before the climber comes to rest. What is the spring constant of the rope, assuming it obeys Hooke's law?

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Webb

Physics

30 January, 13:05

A mountain climber of mass 60.0 kg slips and falls a distance of 4.00 m, at which time he reaches the end of his elastic safety rope. The rope then stretches an additional 2.00 m before the climber comes to rest. What is the spring constant of the rope, assuming it obeys Hooke's law?

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Kinsley Ortiz · Answer 1

The energy that the rope absorbs from the climber is Ep=m*g*h where m is mass of the climber, g=9.81m/s² and h is the height the climber fell. h=4 m+2 m because he was falling for 4 meters and the rope stretched for 2 aditional meters. The potential energy stored in the rope is Er = (1/2) * k*x², where k is the spring constant of the rope and x is the distance the rope stretched and it is

x=2 m. So the equation from the law of conservation of energy is:

Ep=Er

m*g*h = (1/2) * k*x²

k = (2*m*g*h) / x² = (2*60*9.81*6) / 2² = 7063.2/4 = 1765.8 N/m

So the spring constant of the rope is k=1765.8 N/m.