Question

An 86.0-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 1.20 × 10³ N/m. He accidentally slips and falls freely for 0.750 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

Answer:

x = 1.95 m

Explanation:

As we know that total mechanical energy is conserved here

So gravitational potential energy of the climber will convert into spring potential energy

So we will have

$mg(L + x) = \frac{1}{2}kx^2$

here we know that

m = 86 kg

L = 0.750 m

$k = 1.20 \times 10^3 N/m$

now we will have

$86\times 9.8(0.750 + x) = \frac{1}{2}(1.20 \times 10^3) x^2$

$0.750 + x = 0.71 x^2$

$x = 1.95 m$