Question

A mountian climber of mass 60.kg slips and falls distance of 4.0 m, at which time he reaches the end of his elastic saftey rope. The rope then stretches an additional 2.0m before the climber comes to rest. What is the spring constant of the rope , assuming it obeys Hooke's law?

Answer #1

mass of man = 6.0Kg distance of fall= 4.0m stretch of rope= x=2.0m

The loss of potential energy in the fall = m x g x h = 6.0Kg x 9.8 m/sec2 x 4.0 m = 235.2 J

This loss of potential energy appears as Kinetic Energy and kinetic energy gets converted as Elastic potential energy as the rope stretches. Hence the energy 0f 235.2J gets converted to Elastic Potential Energy

Elastic Potential Energy = where k= spring constant and x= extant of stretching of rope

235.2 =

**k** = 2 x 235.2 J /( 2 x 2 mr x mr ) =
**117.2 N/m**

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