A 82.3-kg climber is scaling the vertical wall of a mountain. His safety rope is made of a material that, when stretched, behaves like a spring with a spring constant of 1.73 x 103 N/m. He accidentally slips and falls freely for 0.766 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

1.4521m

Explanation:

The mass of the climber is given as 82.3-kg.

We were told that He accidentally slips and falls freely for 0.766m before the rope runs out of slack wich means that the total when he falls and the rope stretch length is now (0.766m+x).

From the spring formula, the potential energy is equal to the spring force which can be expressed below;

Mgh= ¹/₂kx²

Where m= mass of the climber=82.3-kg

g= acceleration due to gravity= 9.81m/s^2

K= spring constant = 1.73 x 103 N/m

0.5(1700)x² - 82.3(9.81) (0.766+x)

850x² - [807.363(0.766+x)]

= 850x² - 618.44 + 807.363x

= 850x² - 807.363x - 618.44

This is quadratic equation, using the quadratic formula we can get the value of x.

x=- b ±√(b² - 4ac)/2a

x= -(807.363)±√'[(807.363² -(4*859*618.44)]

AFTER SOLVING THE QUADRATIC EQUATION we have

x= -0.501 or 1.4521m

Then we choose the positive value, therefore, the rope stretched with 1.4521m when it breaks his fall and momentarily brings him to rest