# A 95.0-kg climber is scaling the vertical wall of a mountain. His safety rope is made of a...

## Question:

A 95.0 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 {eq}1.25 \times 10^3 \frac{N}{m}{/eq}. He accidentally slips and falls freely for 0.841 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?

## Energy Stored in a Spring:

If we compress or elongate a spring (having spring force constant k) x units beyond its natural length, then the energy stored in the spring or work done on the spring is defined using the following formula:

{eq}W = {E_{stored}} = \dfrac{1}{2}k{x^2} {/eq}

## Answer and Explanation: 1

We are given the following data:

• The mass of the climber is m = 95.0 kg.
• The spring constant is {eq}k = 1.25 \times {10^3}\;{\rm{N/m}} {/eq}
• The distance slips down by the climber (before stretching the rope) is d = 0.841 m.

First, we will write the expression for the change in the gravitational potential energy of the climber:

{eq}\Delta U = mg\left( {d + x} \right) {/eq}, where:

• The elongation in the spring is x.

Writing the expression for the energy stored in the rope (it behaves like a spring) gives us:

{eq}{E_{stored}} = \dfrac{1}{2}k{x^2} {/eq}

We also need to develop an expression for the energy conservation principle:

{eq}\begin{align*} U &= {E_{stored}}\\[0.3cm] mg\left( {d + x} \right) &= \dfrac{1}{2}k{x^2} \end{align*} {/eq}

Substituting the known values into our relations, we have:

{eq}\begin{align*} \left( {95.0\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {0.841\;{\rm{m}} + x} \right) &= \dfrac{1}{2}\left( {1.25 \times {{10}^3}\;{\rm{N/m}}} \right){x^2}\\[0.3cm] 625{x^2} - 783.76995 - 931.95x &= 0\\[0.3cm] 625{x^2} - 931.95x - 783.76995 &= 0\\[0.3cm] x &= \dfrac{{ - \left( {931.95} \right) \pm \sqrt {{{\left( {931.95} \right)}^2} - 4\left( {625} \right)\left( { - 783.76995} \right)} }}{{2\left( {625} \right)}}&\left[\text{Use Quadratic Formula}\right]\\[0.3cm] x &= 0.59976\;{\rm{m}}, - 2.09088\\[0.3cm] x &\approx 0.600\;{\rm{m}} \end{align*} {/eq}

The elongation is positive so we can neglect the negative value of x. Thus, the rope stretches 0.600 meter when it breaks his fall and momentarily brings him to rest.

#### Learn more about this topic: Practice Applying Spring Constant Formulas

from

Chapter 17 / Lesson 11
3.1K

In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.