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Chapter 10: Problem 82

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. \((\mathrm{a})\) 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.

Short Answer

Expert verified
(a) 0.0369 m; (b) 0.0738 m

Step by step solution

01

Understand the Problem

For both parts of the exercise, we need to determine the compression of the spring when a block of mass 0.64 kg is placed on it. Given the spring constant is 170 N/m.
02

Apply Concepts for Part (a)

In part (a), the block rests stationary, which means it is at equilibrium. At equilibrium, the gravitational force on the block equals the spring force. Let the compression of the spring be denoted by \( x_a \). Applying Hooke鈥檚 law and equating forces: \( mg = kx_a \).
03

Calculate Compression for Part (a)

Substitute the known values into the equilibrium equation. \( m = 0.64 \text{ kg}, g = 9.81 \text{ m/s}^2, k = 170 \text{ N/m} \). Thus, \( 0.64 \times 9.81 = 170 \times x_a \). Solving, \( x_a = \frac{0.64 \times 9.81}{170} \approx 0.0369 \text{ m} \).
04

Apply Concepts for Part (b)

In part (b), the block is released and falls, compressing the spring until it momentarily stops. At this point, all gravitational potential energy has been converted into spring potential energy: \( mgh = \frac{1}{2}kx_b^2 \). Here, \( h = x_b \) because the block falls the same distance as the spring's compression.
05

Calculate Compression for Part (b)

Substitute the given values into the energy equation: \( 0.64 \times 9.81 \times x_b = \frac{1}{2} \times 170 \times x_b^2 \). Simplifying, we'll find the root of the equation \( x_b = \frac{2 \times 0.64 \times 9.81}{170} \). Solving gives \( x_b \approx 0.0738 \text{ m} \).
06

Summary of Results

For part (a), the spring compresses by approximately 0.0369 m. For part (b), the spring compresses by approximately 0.0738 m.

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Key Concepts

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

Understanding the Spring Constant
The spring constant, denoted as \( k \), is a fundamental aspect of Hooke's Law, which tells us how stiff a spring is. In our exercise, the spring constant is given as 170 N/m. This value indicates how much force is required to compress or extend the spring by one meter.
  • Higher spring constant = stiffer spring
  • Lower spring constant = looser spring
In mathematical terms, Hooke's Law is represented as \( F = kx \), where \( F \) is the force applied to the spring, \( k \) is the spring constant, and \( x \) is the displacement or compression distance. In our problem, solving for \( x \) helps determine how much the spring is compressed by the block.
Finding the Equilibrium Position
The equilibrium position is the point at which the forces acting on an object are balanced, resulting in no net movement. In the context of our exercise, when the block rests on the spring without any motion, it means it has reached its equilibrium position.
  • Gravitational force \((mg)\) pulls the block downward.
  • Spring force \((kx)\) pushes upward, equalizing the gravitational pull.
At this position, the compression of the spring can be found using Hooke's Law. By equating the gravitational force and the spring force, the displacement \( x \) can be solved, which in our case, results in a compression of approximately 0.0369 m for Part (a).
Exploring Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. When an object is lifted, it gains potential energy, which can later be converted into kinetic or other forms of energy as it falls.
In our exercise, when the block is released and falls onto the spring, its gravitational potential energy is converted into the energy stored in the spring. GPE is calculated using the formula \( mgh \), where \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is the height or distance fallen, which is equal to the spring's compression in this context.
Understanding Spring Potential Energy
Spring potential energy is the energy stored within a spring when it is compressed or stretched. This potential energy is determined by the formula \( \frac{1}{2} kx^2 \), where \( k \) is the spring constant and \( x \) is the displacement.
  • When compressed, the spring stores more energy.
  • The energy can be transferred once the spring is released.
In Part (b) of our exercise, the potential energy within the spring after being compressed must equal the gravitational potential energy lost by the block. By setting \( mgh = \frac{1}{2}kx^2 \), we can solve for \( x \) to find the maximum compression of approximately 0.0738 m.

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Most popular questions from this chapter

Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is 174 \(\mathrm{N} / \mathrm{m}\) . The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2 . The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2 .

In a room that is 2.44 m high, a spring (unstrained length \(=0.30 \mathrm{m} )\) hangs from the ceiling. A board whose length is 1.98 \(\mathrm{m}\) is attached to the free end of the spring. The board hangs straight down, so that its \(1.98-\mathrm{m}\) length is perpendicular to the floor. The weight of the board (104 \(\mathrm{N} )\) stretches the spring so that the lower end of the board just extends to, but does not touch, the floor. What is the spring constant of the spring?

A spring is resting vertically on a table. A small box is dropped onto the top of the spring and compresses it. Suppose the spring has a spring constant of 450 \(\mathrm{N} / \mathrm{m}\) and the box has a mass of 1.5 \(\mathrm{kg}\) . The speed of the box just before it makes contact with the spring is 0.49 \(\mathrm{m} / \mathrm{s}\) . (a) Determine the magnitude of the spring's displacement at an instant when the acceleration of the box is zero. (b) What is the magnitude of the spring's displacement when the spring is fully compressed?

\(\mathrm{A} 1.00 \times 10^{-2}\) -kg bullet is fired horizontally into a \(2.50-\mathrm{kg}\) wooden block attached to one end of a massless horizontal spring \((k=845 \mathrm{N} / \mathrm{m})\) The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.200 \(\mathrm{m}\) . What is the speed of the bullet?

A spring is hung from the ceiling. A \(0.450-\mathrm{kg}\) block is then attached to the free end of the spring. When released from rest, the block drops 0.150 \(\mathrm{m}\) before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.
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