/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 An 86.0 -kg climber is scaling t... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(1.20 \times 10^{3} \mathrm{~N} / \mathrm{m} .\) He accidentally slips and falls freely for \(0.750 \mathrm{~m}\) before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

Short Answer

Expert verified
The rope stretches approximately 1.027 meters.

Step by step solution

01

Identify Forces and Set up the Problem

The climber falls and his potential energy is converted into kinetic energy. When the rope stretches to bring him to rest, his kinetic energy is converted into the elastic potential energy stored in the rope.
02

Calculate Initial Potential Energy

The initial potential energy (PE) of the climber due to gravity can be calculated using the formula:\[ \text{PE} = mgh \]where,\( m = 86.0 \, \text{kg}\),\( g = 9.81 \, \text{m/s}^2 \),\( h = 0.750 \, \text{m} \).Substitute these values:\[ \text{PE} = 86.0 \times 9.81 \times 0.750 = 632.895 \, \text{J} \].
03

Equate Kinetic Energy to Elastic Potential Energy

The kinetic energy gained by the climber is transformed into elastic potential energy when the rope stretches. The elastic potential energy stored in the rope can be expressed as:\[ \text{Elastic PE} = \frac{1}{2} k x^2 \] where \( k = 1.20 \times 10^3 \, \text{N/m} \) is the spring constant and \( x \) is the stretch in the rope.Set the potential energy equal to the elastic potential energy:\[ 632.895 = \frac{1}{2} \times 1.20 \times 10^3 \times x^2 \].
04

Solve for Rope Stretch

Rearrange the equation from the previous step and solve for \( x \):\[ 632.895 = 600x^2 \]\[ x^2 = \frac{632.895}{600} \]\[ x^2 \approx 1.054825 \]\[ x \approx \sqrt{1.054825} \approx 1.027 \, \text{m} \]. Thus, the rope stretches approximately \(1.027\, \text{m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Constant
The spring constant, often denoted as \( k \), is a measure of a spring's stiffness. A larger \( k \) value indicates a stiffer spring, which requires more force to stretch. In the context of our climber, his safety rope acts like a spring with a spring constant of \( 1.20 \times 10^{3} \, \text{N/m} \). This means that every meter the rope is stretched, it exerts a force of approximately \( 1200 \, \text{N} \) to return to its original length.

Understanding the spring constant helps in determining how far a spring (or the rope) can stretch when a force (such as the climber's fall) is applied to it:
  • The spring constant is crucial in calculating elastic potential energy.
  • It helps predict how much energy is stored when the rope stretches.
  • Thus, it is vital for the safety design of ropes in mountaineering.
In any situation where a spring-like object is used to absorb energy, knowing the spring constant allows engineers to ensure the object functions safely and effectively.
Potential Energy
Potential energy is the stored energy possessed by an object due to its position or state. In our exercise, the climber initially has gravitational potential energy when he is 0.75 meters above where the rope will stop him. This energy can be calculated using the formula \( \text{PE} = mgh \). Here, \( m \) is the climber's mass, \( g \) is the acceleration due to gravity, and \( h \) is the height fallen.

In this scenario, the potential energy is a key contributor to:
  • The kinetic energy the climber acquires as he falls.
  • The energy that eventually converts into the rope's elastic potential energy.
  • Understanding how energy is conserved and transferred in mechanical systems.
As the climber falls, his potential energy decreases, while his kinetic energy increases equivalently, maintaining energy conservation throughout his motion until the rope stretches.
Kinetic Energy
Kinetic energy is the energy of motion. When the climber falls freely, his potential energy due to height is converted into kinetic energy due to speed. The formula for kinetic energy is \( \text{KE} = \frac{1}{2} mv^2 \), but in this exercise, we do not need to calculate it directly; we instead know all this kinetic energy transforms into elastic potential energy as the rope stretches.

Here's how kinetic energy plays a role in the fall:
  • It is the bridge between potential energy and the elastic potential energy of the rope.
  • At the peak of his fall, all the climber's energy is in the form of kinetic energy.
  • Understanding kinetic energy helps in designing systems that can safely absorb and dissipate energy, like shock absorbers or safety ropes.
Recognizing that energy transformations don't lose the quantity of energy involved is crucial in understanding how forces do work, and how safety systems should be engineered.

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Most popular questions from this chapter

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of \(f=2.00 \mathrm{~Hz}\). On this tray is an empty cup. Obtain the coefficient of static friction between the tray and the cup, given that the cup begins slipping when the amplitude of the motion is \(5.00 \times 10^{-2} \mathrm{~m}\).

Concept Simulation 10.2 at allows you to explore the effect of the acceleration due to gravity on pendulum motion, which is the focus of this problem. Astronauts on a distant planet set up a simple pendulum of length \(1.2 \mathrm{~m}\). The pendulum executes simple harmonic motion and makes 100 complete vibrations in \(280 \mathrm{~s}\). What is the acceleration due to gravity?

Refer to Interactive Solution \(\underline{10.77}\) at to review a method by which this problem can be solved. An \(11.2-\mathrm{kg}\) block and a \(21.7-\mathrm{kg}\) block are resting on a horizontal frictionless surface. Between the two is squeezed a spring (spring constant \(=1330 \mathrm{~N} / \mathrm{m}\) ). The spring is compressed by \(0.141 \mathrm{~m}\) from its unstrained length and is not attached permanently to either block. With what speed does each block move away after the mechanism keeping the spring squeezed is released and the spring falls away?

A die is designed to punch holes with a radius of \(1.00 \times 10^{-2} \mathrm{~m}\) in a metal sheet that is \(3.0 \times 10^{-3} \mathrm{~m}\) thick, as the drawing illustrates. To punch through the sheet, the die must exert a shearing stress of \(3.5 \times 10^{8} \mathrm{~Pa}\). What force \(\overrightarrow{\mathbf{F}}\) must be applied to the die?

Interactive Solution \(\underline{10.21}\) at presents a model for solving this problem. A spring (spring constant \(=112 \mathrm{~N} / \mathrm{m}\) ) is mounted on the floor and is oriented vertically. A 0.400 kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in \(\mathrm{Hz}\) ) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

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