/*! 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 112 A \(70.0 \mathrm{~kg}\) man jump... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(70.0 \mathrm{~kg}\) man jumping from a window lands in an elevated fire rescue net \(11.0 \mathrm{~m}\) below the window. He momentarily stops when he has stretched the net by \(1.50 \mathrm{~m}\). Assuming that mechanical energy is conserved during this process and that the net functions like an ideal spring, find the elastic potential energy of the net when it is stretched by \(1.50 \mathrm{~m}\).

Short Answer

Expert verified
The elastic potential energy of the net is 8820.15 J.

Step by step solution

01

Understand the Problem

A man jumps from a window and lands in a net, which behaves like an ideal spring. We are to find the elastic potential energy of the net when the net is stretched by 1.50 m, given that mechanical energy is conserved.
02

Define Known Quantities and Relations

The man's mass is \(70.0 \text{ kg}\) and he falls 11.0 m before coming to a stop after stretching the net by 1.50 m. Since the net acts like a spring, Hooke's Law applies: \(U = \frac{1}{2} k x^2\), where \(U\) is the elastic potential energy, \(k\) is the spring constant, and \(x = 1.50 \text{ m}\). Also, the gravitational potential energy initially lost is \(mgh\).
03

Calculate Initial Gravitational Potential Energy

Calculate the gravitational potential energy lost: \[ PE_{initial} = mgh = 70.0 \text{ kg} \times 9.81 \text{ m/s}^2 \times (11.0 \text{ m} + 1.50 \text{ m}) = 8820.15 \text{ J} \]
04

Set Energy Conservation Equation

According to energy conservation, the gravitational potential energy lost equals the elastic potential energy gained by the net at maximum extension. Thus, \[ PE_{initial} = U \] This gives: \[ U = 8820.15 \text{ J} \]
05

Conclude the Solution

The elastic potential energy of the net when the man momentarily stops after stretching it by 1.50 m is 8820.15 J.

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Key Concepts

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

Mechanical Energy Conservation
Mechanical energy conservation is a fundamental concept in physics stating that the total mechanical energy of a system remains constant if only conservative forces, like gravity and elasticity, act upon it. This principle applies to scenarios like a man jumping into a net that stretches like a spring. In this case, the mechanical energy in the system is made up of gravitational potential energy and elastic potential energy.
  • Gravitational potential energy is the energy stored due to an object's position relative to Earth's surface.
  • Elastic potential energy is stored when an object is stretched or compressed, such as a spring.
When the man falls, his gravitational potential energy is converted into elastic potential energy in the net. The initial potential energy, calculated from the height he fell from, is transformed entirely into the elastic energy at the point of maximum stretch. Hence, mechanical energy conservation helps us understand the relationship between the man's fall and the net's stretch. It explains why the energy transferred by his fall equals the energy stored in the net.
Hooke's Law
Hooke's Law is a principle that describes the behavior of springs and other elastic materials. It states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, it is represented as:
\[ F = kx \] where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the original length.
In the given scenario, the net functions as an ideal spring. Hooke's Law helps us determine how the gravitational potential energy of the man converts into elastic potential energy as the net stretches. The spring constant \( k \) relates the stretch of the net to the force exerted by it.
The energy stored due to the stretch, known as elastic potential energy, is given by:
\[ U = \frac{1}{2} k x^2 \] where \( U \) is the elastic potential energy of the net, calculated at the point of maximum stretch. Understanding Hooke's Law allows us to deduce how much energy is stored in the stretched net based on how much it stretches and its inherent spring constant.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field, typically related to its height above the ground. This energy is determined using the formula:
\[ PE = mgh \] where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( h \) is the height.
In our example, the man starts with a certain amount of gravitational potential energy at the height of 11.0 meters above the net. As he falls, this potential energy gets converted into other forms of mechanical energy—in this case, it becomes elastic potential energy when the net stretches. Determining the initial gravitational potential energy allows us to understand how much energy will be converted to elastic potential energy in the net. This conversion process illuminates the conservation of mechanical energy, where potential energy at a height equals the potential energy stored in the stretched net.

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Most popular questions from this chapter

A spring with spring constant \(k=620 \mathrm{~N} / \mathrm{m}\) is placed in a vertical orientation with its lower end supported by a horizontal surface. The upper end is depressed \(25 \mathrm{~cm},\) and a block with a weight of \(50 \mathrm{~N}\) is placed (unattached) on the depressed spring. The system is then released from rest. Assume that the gravitational potential energy \(U_{g}\) of the block is zero at the release point \((y=0)\) and calculate the kinetic energy \(K\) of the block for \(y\) equal to \((\mathrm{a}) 0,(\mathrm{~b}) 0.050 \mathrm{~m},(\mathrm{c}) 0.10 \mathrm{~m},(\mathrm{~d}) 0.15 \mathrm{~m},\) and \((\mathrm{e}) 0.20 \mathrm{~m} .\) Also (f) how far above its point of release does the block rise?

A spring \((k=200 \mathrm{~N} / \mathrm{m})\) is fixed at the top of a frictionless plane inclined at angle \(\theta=40^{\circ}(\) Fig. \(8-59) .\) A \(1.0 \mathrm{~kg}\) block is projected up the plane, from an initial position that is distance \(d=0.60 \mathrm{~m}\) from the end of the relaxed spring, with an initial kinetic energy of 16 J. (a) What is the kinetic energy of the block at the instant it has compressed the spring \(0.20 \mathrm{~m} ?\) (b) With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by \(0.40 \mathrm{~m} ?\)

A \(5.0 \mathrm{~g}\) marble is fired vertically upward using a spring gun. The spring must be compressed \(8.0 \mathrm{~cm}\) if the marble is to just reach a target \(20 \mathrm{~m}\) above the marble's position on the compressed spring. (a) What is the change \(\Delta U_{g}\) in the gravitational potential energy of the marble-Earth system during the \(20 \mathrm{~m}\) ascent? (b) What is the change \(\Delta U_{s}\) in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?

Next, the particle is released from rest at \(x=0 .\) What are (l) its kinetic energy at \(x=5.0 \mathrm{~m}\) and \((\mathrm{m})\) the maximum positive position \(x_{\max }\) it reaches? (n) What does the particle do after it reaches \(x_{\max } ?\) $$ \begin{array}{lc} {\text { Range }} & \text { Force } \\ \hline 0 \text { to } 2.00 \mathrm{~m} & \vec{F}_{1}=+(3.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 2.00 \mathrm{~m} \text { to } 3.00 \mathrm{~m} & \vec{F}_{2}=+(5.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 3.00 \mathrm{~m} \text { to } 8.00 \mathrm{~m} & F=0 \\ 8.00 \mathrm{~m} \text { to } 11.0 \mathrm{~m} & \vec{F}_{3}=-(4.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 11.0 \mathrm{~m} \text { to } 12.0 \mathrm{~m} & \vec{F}_{4}=-(1.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 12.0 \mathrm{~m} \text { to } 15.0 \mathrm{~m} & F=0 \end{array} $$ For the arrangement of forces in Problem \(81,\) a \(2.00 \mathrm{~kg}\) particle is released at \(x=5.00 \mathrm{~m}\) with an initial velocity of \(3.45 \mathrm{~m} / \mathrm{s}\) in the negative direction of the \(x\) axis. (a) If the particle can reach \(x=0 \mathrm{~m},\) what is its speed there, and if it cannot, what is its turning point? Suppose, instead, the particle is headed in the positive \(x\) direction when it is released at \(x=5.00 \mathrm{~m}\) at speed \(3.45 \mathrm{~m} / \mathrm{s} .\) (b) If the particle can reach \(x=13.0 \mathrm{~m},\) what is its speed there, and if it cannot, what is its turning point?

An outfielder throws a baseball with an initial speed of \(81.8 \mathrm{mi} / \mathrm{h} .\) Just before an infielder catches the ball at the same level, the ball's speed is \(110 \mathrm{ft} / \mathrm{s}\). In foot-pounds, by how much is the mechanical energy of the ball-Earth system reduced because of air drag? (The weight of a baseball is 9.0 oz.)

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