Question

An 85.4-kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of 1.19 103 N/m. He accidentally slips and falls freely for 0.815 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

Answer 1

This is a conservation of energy problem.

One balances the potential energy lost by the falling climber with the ``spring'' energy in the rope.

He falls from rest and comes to rest at the end so kinetic energy will be zero both initially and finally. The only subtle point is that the climber continues to lose potential energy as the rope is breaking his fall.

Let be the amount the rope stretches.

Measuring the potential energy from the final stretched position of the rope, one finds that the initial energy is ( is the distance he fell before the rope tightened).

Then the final energy is purely the energy stored in the spring

Now,

Equating the initial and final energies one gets:

Plugging in the values, we get:

On solving, we get

x = 1.98 m

Hence, the rope is stretched by 1.98 m