# 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}

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.

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.