/*! 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 34 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 to break the fall.

Step by step solution

01

Determine Potential Energy Before Fall

The potential energy lost by the climber when falling must be converted into elastic energy stored in the rope. Calculate the gravitational potential energy (PE) lost using the formula: \[ PE = mgh \]where \(m = 86.0 \text{ kg}\), \(g = 9.81 \text{ m/s}^2\), and \(h = 0.750 \text{ m}\). Substitute these values:\[ PE = 86.0 \times 9.81 \times 0.750 = 632.355 \text{ J} \]
02

Relate Elastic Potential Energy to Solar Work

When the climber is momentarily at rest, all the potential energy he lost is stored as elastic potential energy in the rope. Elastic potential energy in the spring (or elastic rope in this context) is given by:\[ U = \frac{1}{2} k x^2 \]where \(k\) is the spring constant \(1.20 \times 10^3 \text{ N/m}\) and \(x\) is the amount of stretch in the rope we need to find.
03

Equate Potential Energy to Elastic Energy

Set the potential energy equal to the elastic potential energy:\[ 632.355 = \frac{1}{2} \times 1.20 \times 10^3 \times x^2 \] This simplifies to:\[ x^2 = \frac{2 \times 632.355}{1.20 \times 10^3} = \frac{1264.71}{1200} \]
04

Solve for the Stretch of the Rope

Compute the value of \(x^2\) from the previous step:\[ x^2 = 1.053925 \] Take the square root of both sides to solve for \(x\):\[ x = \sqrt{1.053925} \approx 1.027 \text{ m} \]
05

Verify the Answer

Finally, verify your solution by plugging \(x\) back into the elastic potential energy formula and checking it matches the potential energy:\[ U = \frac{1}{2} \times 1.20 \times 10^3 \times (1.027)^2 = 632.355 \text{ J} \]Thus, the calculations confirm that 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.

Gravitational Potential Energy
In the context of the exercise, gravitational potential energy plays a crucial role. It is the energy stored because of an object's position relative to Earth. For the climber, this energy is significant owing to his height above the ground.
Gravitational potential energy is determined using the formula:
  • \(PE = mgh\)
  • \(m\) is the mass of the object (in kilograms)
  • \(g\) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\) on Earth)
  • \(h\) is the height above the ground (in meters)
In the climber's case, falling \(0.750 \text{ m}\) means his potential energy decreases, converting into another energy form.
This energy conversion is crucial to understanding how and why the rope stretches when it stops his fall.
Spring Constant
The spring constant is a measure of the stiffness of a spring. In this exercise, the nylon rope has spring-like qualities, described by its spring constant. It is represented as \(k\) and expressed in Newtons per meter (\(\text{N/m}\)).
A higher spring constant means a stiffer spring that doesn't stretch as easily. For the climber's rope, we have:
  • Spring constant \(k = 1.20 \times 10^3 \text{ N/m}\)
In the realm of elastic potential energy, the spring constant combines with the distance the spring (or in this case, the rope) stretches. This relationship determines how much energy is stored in the stretched spring. The formula for elastic potential energy is:
  • \(U = \frac{1}{2} k x^2\)
  • \(x\) is the displacement or stretch of the spring (in meters)
The rope's ability to absorb energy and prevent the climber from falling further is directly related to its spring constant.
Mechanical Energy Conservation
Mechanical energy conservation is a principle that states the total mechanical energy in a closed system remains constant as long as only conservative forces (like gravity and spring force) are involved.
In this problem, when the climber falls, his potential energy is converted to kinetic energy until the rope stretches. At this point, kinetic energy is exchanged for elastic potential energy.
Conservation of mechanical energy means:
  • Potential energy lost due to the fall becomes elastic potential energy.
This can be expressed as:
  • \(mgh = \frac{1}{2}kx^2\)
The equation shows that the gravitational potential energy initially present is equal to the elastic potential energy stored when the rope fully stretches. This framework aids in predicting energy transfers that occur during the climber's fall and subsequent stop.
Nylon Rope Elasticity
Nylon rope elasticity refers to the rope's ability to stretch and absorb energy without breaking. It acts like a spring in energy conversion scenarios, where its elasticity can help to soften and safely halt falls.
The elasticity of the rope is vital in ensuring the climber comes to rest without experiencing abrupt deceleration forces, which could cause injury. The way the rope behaves is summarized as:
  • The rope stretches as it absorbs energy.
  • The spring constant denotes the rope's stiffness.
In practical terms, this elasticity provides a cushioning effect during a fall, reducing the impact force on the climber. The rope behaves similarly to an ideal spring, storing elastic potential energy which is calculated using the formula \(U = \frac{1}{2} k x^2\).
Nylon's material properties, including its density and flexibility, make it suitable for use in climbing ropes where both strength and reasonable elasticity are desired.

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Most popular questions from this chapter

In 0.750 s, a 7.00-kg block is pulled through a distance of 4.00 m on a frictionless horizontal surface, starting from rest. The block has a constant acceleration and is pulled by means of a horizontal spring that is attached to the block. The spring constant of the spring is 415 N/m. By how much does the spring stretch?

In preparation for shooting a ball in a pinball machine, a spring \((k=675 \mathrm{N} / \mathrm{m})\) is compressed by \(0.0650 \mathrm{m}\) relative to its unstrained length. The ball \((m=0.0585 \mathrm{kg})\) is at rest against the spring at point A. When the spring is released, the ball slides (without rolling). It leaves the spring and arrives at point \(\mathrm{B},\) which is \(0.300 \mathrm{m}\) higher than point A. Ignore friction, and find the ball's speed at point B.

A simple pendulum is made from a 0.65-m-long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released, how much time elapses before it attains its greatest speed?

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?

\(\mathrm{A}\) square plate is \(1.0 \times 10^{-2} \mathrm{m}\) thick, measures \(3.0 \times 10^{-2} \mathrm{m}\) on a side, and has a mass of \(7.2 \times 10^{-2} \mathrm{kg} .\) The shear modulus of the material is \(2.0 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) One of the square faces rests on a flat horizontal surface, and the coefficient of static friction between the plate and the surface is 0.91. A force is applied to the top of the plate, as in Figure 10.29 a. Determine (a) the maximum possible amount of shear stress, (b) the maximum possible amount of shear strain, and (c) the maximum possible amount of shear deformation \(\Delta X\) (see Figure \(10.29 b\) ) that can be created by the applied force just before the plate begins to move.

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