/*! 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] | 91Ó°ÊÓ

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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
4.8 cm above the spring.

Step by step solution

01

Understanding the Problem

To solve this problem, we need to determine the height from which the block was dropped. The potential energy of the block at the height equals the elastic potential energy when the block compresses the spring.
02

Converting Units

The spring compression is given in centimeters, so we need to convert it to meters for consistency in our calculations with other SI units. Therefore, the compression is \[x = 2.5 \text{ cm} = 0.025 \text{ m}. \]
03

Identifying the Energy Transformation

Initially, the block has gravitational potential energy which is converted into the spring's elastic potential energy when the block compresses the spring. We use the formula for gravitational potential energy \[U_g = m g h,\]and for elastic potential energy \[U_e = \frac{1}{2} k x^2.\]
04

Setting Potential Energies Equal

At the moment the spring is fully compressed, the gravitational potential energy is completely converted into elastic potential energy. Therefore, set \[m g h = \frac{1}{2} k x^2,\]and solve for the height \( h \).
05

Solving for Height

Substitute the known values into the equation. We have:\[0.30 \cdot 9.8 \cdot h = \frac{1}{2} \cdot 450 \cdot (0.025)^2.\]Simplify and solve for \( h \):\[0.3 \cdot 9.8 \cdot h = \frac{1}{2} \cdot 450 \cdot 0.000625,\]\[2.94h = 0.140625,\]\[h = \frac{0.140625}{2.94} \approx 0.048 \text{ meters}.\]
06

Converting Height to Centimeters

Convert the height back to centimeters to answer the question in the required unit:\[h = 0.048 \text{ m} \times 100 \frac{\text{cm}}{\text{m}} = 4.8 \text{ cm}.\]

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Key Concepts

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

Gravitational Potential Energy
Gravitational Potential Energy is the energy stored in an object due to its position in a gravitational field. When you lift an object off the ground, you give it gravitational potential energy. The energy is determined by the object's mass, the height it is raised, and the strength of the gravitational field. The formula for gravitational potential energy is \[U_g = mgh\],where:
  • \(U_g\) is the gravitational potential energy.
  • \(m\) is the mass of the object in kilograms.
  • \(g\) is the acceleration due to gravity, approximately \(9.8 \, \text{m/s}^2\) on Earth.
  • \(h\) is the height in meters.
This energy converts to other forms when the object moves, such as when a block falls toward a spring. As it falls, gravitational potential energy decreases, transforming into other energy forms like kinetic energy or elastic potential energy once it hits and compresses a spring.
Understanding this concept helps in analyzing how energies transform in physical systems, like in this case, where the energy at height becomes the energy that compresses a spring.
Elastic Potential Energy
Elastic Potential Energy is the energy stored in elastic materials as a result of their stretching or compressing. When an object like a spring is compressed or stretched, it stores energy that can be released later. The energy depends on both the amount of stretch/compression and the spring’s stiffness, known as the spring constant.
The formula for calculating elastic potential energy is:\[U_e = \frac{1}{2} k x^2\]where:
  • \(U_e\) is the elastic potential energy.
  • \(k\) is the spring constant, which indicates how stiff the spring is.
  • \(x\) is the displacement from the equilibrium position.
In our exercise, when the block hits the spring and compresses it, the gravitational potential energy converts into elastic potential energy. This stored energy is what slows down the block, bringing it momentarily to rest. Understanding this concept allows one to see how systems store and release energy, such as springs in clocks or shock absorbers in vehicles.
Spring Constant
The Spring Constant is a measure of a spring's stiffness. It provides an indication of how much force you need to apply to compress or stretch the spring by a certain amount. Denoted by \(k\), it is expressed in units of Newtons per meter (\(\text{N/m}\)).
In Hooke's Law, the spring constant relates the force exerted on the spring and the displacement it causes:\[F = kx\]where:
  • \(F\) is the force applied to the spring.
  • \(k\) is the spring constant.
  • \(x\) is the displacement of the spring from its initial position.
A larger spring constant means a stiffer spring, requiring more force for the same amount of displacement. For instance, in this exercise, a spring constant of 450 N/m means the spring is relatively stiff, requiring a significant force for compression. Understanding the spring constant is crucial for calculating elastic potential energy and interpreting how physical systems behave, especially in cases involving energy transformations like the one in our block and spring scenario.

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Most popular questions from this chapter

55m A person who weighs \(670 \mathrm{N}\) steps onto a spring scale in the bathroom, and the spring compresses by \(0.79 \mathrm{cm} .\) (a) What is the spring constant? (b) What is the weight of another person who compresses the spring by \(0.34 \mathrm{cm} ?\)

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is \(f=3.0 \mathrm{Hz} ?\)

When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amplitude of \(6.3 \times 10^{-7}\) m and a maximum speed of \(2.9 \times 10^{-3} \mathrm{m} / \mathrm{s}\) (a) What is the frequency (in Hz) of the eardrum’s vibration? (b) What is the maximum acceleration of the eardrum?

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?

Two metal beams are joined together by four rivets, as the drawing indicates. Each rivet has a radius of \(5.0 \times 10^{-3} \mathrm{m}\) and is to be exposed to a shearing stress of no more than \(5.0 \times 10^{8}\) Pa. What is the maximum tension \(\overrightarrow{\mathbf{T}}\) that can be applied to each beam, assuming that each rivet carries one-fourth of the total load?

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