/*! 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 29 A vertical spring with a spring ... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 29

A vertical spring with a spring constant of \(450 \mathrm{N} / \mathrm{m}\) is mounted on the floor. From directly above the spring, which is unstrained, a \(0.30-\mathrm{kg}\) block is dropped from rest. It collides with and sticks to the spring, which is compressed by \(2.5 \mathrm{cm}\) in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in \(\mathrm{cm}\) ) above the compressed spring was the block dropped?

Short Answer

Expert verified
The block was dropped from approximately 4.784 cm above the spring.

Step by step solution

01

Understand the Energy Transformation

The potential energy (PE) of the block is transformed into elastic potential energy (EPE) of the spring when the block compresses the spring. Since the block is initially dropped from some height, its initial PE is given by \( PE = mgh \). As it compresses the spring, this energy gets converted into the spring's elastic potential energy, \( EPE = \frac{1}{2}kx^2 \), where \( k \) is the spring constant and \( x \) is the compression of the spring.
02

Set Up the Conservation of Energy Equation

We set the initial potential energy \( mgh \) equal to the elastic potential energy \( \frac{1}{2}kx^2 \) at the point where the spring is fully compressed and the block momentarily stops. Therefore, the equation is: \[ mgh = \frac{1}{2}kx^2 \]
03

Plug in the Known Values

Plugging in the known values: - \( m = 0.30 \text{ kg} \)- \( g = 9.8 \text{ m/s}^2 \)- \( k = 450 \text{ N/m} \)- \( x = 0.025 \text{ m} \) (converted from 2.5 cm)The equation becomes: \[ 0.30 \cdot 9.8 \cdot h = \frac{1}{2} \cdot 450 \cdot 0.025^2 \]
04

Solve for Height \(h\)

Simplifying the right-hand side of the equation gives:\[ \frac{1}{2} \cdot 450 \cdot 0.025^2 = 0.140625 \] So the equation is:\[ 2.94h = 0.140625 \]Solving for \( h \), we get:\[ h = \frac{0.140625}{2.94} \approx 0.04784 \text{ m} \]
05

Convert Meters to Centimeters

Since the problem asks for the height in centimeters, convert meters to centimeters:\[ 0.04784 \text{ m} = 4.784 \text{ cm} \]

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Key Concepts

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

Elastic Potential Energy
Elastic potential energy is the energy stored in an object when it is stretched or compressed. Springs are a common example where this type of energy is involved.
The formula for calculating elastic potential energy is given by:
  • \[ EPE = \frac{1}{2} k x^2 \]
Where:
  • \( EPE \) is the elastic potential energy.
  • \( k \) is the spring constant, a measure of the stiffness of the spring.
  • \( x \) is the compression or stretch distance from the spring's equilibrium position.
When a block falls on the spring and compresses it, the gravitational potential energy of the block is converted into elastic potential energy. This energy transformation follows the conservation of energy principle.
Spring Constant
The spring constant \( k \) is a measure of a spring's stiffness. It tells us how much force is needed to compress or stretch the spring by a unit length. The unit for spring constant is Newtons per meter (N/m).
If you have a high spring constant, the spring is very stiff and requires a lot of force to change its shape. Conversely, a spring with a low constant is more flexible and compresses easily with less force.
For example, in the exercise, the spring constant is \( 450 \mathrm{N/m} \), indicating a relatively stiff spring. This stiffness dictates how much energy can be stored as elastic potential energy when the spring is compressed by a certain amount.
Gravitational Potential Energy
Gravitational potential energy (GPE) describes the energy an object has due to its position in a gravitational field. When an object is lifted to a height, it gains gravitational potential energy. The formula to calculate GPE is:
  • \[ PE = mgh \]
Where:
  • \( PE \) is the potential energy.
  • \( m \) is the mass of the object.
  • \( g \) is the acceleration due to gravity (usually \( 9.8 \, \text{m/s}^2 \) on Earth).
  • \( h \) is the height above a reference point.
In the example, the block is initially held above the spring, converting its gravitational potential energy into elastic potential energy when dropped. This conversion demonstrates the principle of conservation of energy, where energy changes forms but is not lost.

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Most popular questions from this chapter

A vertical ideal spring is mounted on the floor and has a spring constant of \(170 \mathrm{N} / \mathrm{m} .\) A \(0.64-\mathrm{kg}\) block is placed on the spring in two different ways. (a) In one case, the block is placed on the spring and not released until it rests stationary on the spring in its equilibrium position. Determine the amount (magnitude only) by which the spring is compressed. (b) In a second situation, the block is released from rest immediately after being placed on the spring and falls downward until it comes to a momentary halt. Determine the amount (magnitude only) by which the spring is now compressed.

A loudspeaker diaphragm is producing a sound for 2.5 s by moving back and forth in simple harmonic motion. The angular frequency of the motion is \(7.54 \times 10^{4} \mathrm{rad} / \mathrm{s} .\) How many times does the diaphragm move back and forth?

A person bounces up and down on a trampoline, while always staying in contact with it. The motion is simple harmonic motion, and it takes 1.90 s to complete one cycle. The height of each bounce above the equilibrium position is \(45.0 \mathrm{cm} .\) Determine (a) the amplitude and (b) the angular frequency of the motion. (c) What is the maximum speed attained by the person?

A \(1.00 \times 10^{-2}\) -kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is \(124 \mathrm{N} / \mathrm{m}\). The block is shoved parallel to the spring axis and is given an initial speed of \(8.00 \mathrm{m} / \mathrm{s},\) while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

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 \(\mathrm{A} .\) 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 \(\mathrm{A} .\) Ignore friction, and find the ball's speed at point \(\mathrm{B}\).
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