/*! 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 36 A spring is resting vertically o... [FREE SOLUTION] | 91Ó°ÊÓ

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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 N/m and the box has a mass of 1.5 kg. The speed of the box just before it makes contact with the spring is 0.49 m/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?

Short Answer

Expert verified
(a) 0.0327 m (b) 0.0159 m

Step by step solution

01

Understand the problem

We're solving a problem involving a box compressing a spring. We need to calculate the displacement of the spring at two different states: when the acceleration of the box is zero and when the spring is fully compressed.
02

Identify the relevant physics concepts

At the point when the acceleration of the box is zero, the forces on the box are balanced. Therefore, the gravitational force on the box is balanced by the spring force. When fully compressed, the box is momentarily at rest, so the mechanical energy is conserved.
03

Calculate the displacement when acceleration is zero

The forces balancing condition means that the gravitational force is equal to the spring force: \[ mg = kx \]where: \( m = 1.5 \) kg (mass of the box), \( g = 9.81 \) m/sty2 (acceleration due to gravity), and \( k = 450 \) N/m (spring constant). Solve for \( x \): \[ x = \frac{mg}{k} = \frac{1.5 \times 9.81}{450} \approx 0.0327 \text{ m} \]
04

Calculate the maximum compression of the spring

At maximum compression, the kinetic energy is converted into potential energy stored within the spring.The potential energy of the compressed spring is given by: \[ rac{1}{2} k x^2 = rac{1}{2} m v^2 \]where \( v = 0.49 \) m/s is the initial speed of the box.Rearranging and solving gives:\[ x^2 = \frac{m v^2}{k} = \frac{1.5 \times 0.49^2}{450} \]Calculate for \( x \): \[ x = \sqrt{\frac{1.5 \times 0.49^2}{450}} \approx 0.0159 \text{ m} \]

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Key Concepts

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

spring constant
Every spring has a specific stiffness, measured by a value known as the spring constant, denoted as \( k \). It represents the force required to displace the spring by a unit distance. The unit for the spring constant is Newtons per meter (N/m). The higher the spring constant, the stiffer the spring, meaning it requires more force to compress or stretch it by a specific amount. For example, in our exercise, the spring constant is 450 N/m, which means a force of 450 N is needed to compress it by 1 meter.
mechanical energy conservation
In physics, mechanical energy conservation is an important principle stating that the total mechanical energy of a system remains constant if the forces acting are conservative. This means that energy can transform from one form to another without being lost. In a spring system, potential energy and kinetic energy are the two main forms involved. When the box hits the spring and comes to rest momentarily, its entire initial kinetic energy is converted into potential energy stored in the spring. This is described by the equation \( \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \), where \( k \) is the spring constant, \( x \) is the spring displacement, \( m \) is the mass of the object, and \( v \) is its velocity.
forces in equilibrium
When a system is in equilibrium, all the forces acting upon it are balanced, resulting in a state of no net force and no acceleration. In the context of our spring problem, the box's acceleration is zero when the spring force equals the gravitational force acting on the box. This condition can be mathematically represented as \( mg = kx \), where \( m \) is the mass of the box, \( g \) is the acceleration due to gravity, and \( x \) is the displacement in this balanced state. Understanding equilibrium helps in determining situations where forces cancel each other out, allowing us to calculate specific displacements.
spring displacement
Spring displacement refers to the distance a spring is compressed or extended from its natural, undisturbed position. In our exercise, we calculate displacement at two crucial points: when the acceleration of the box is zero and when the spring is fully compressed. Using the formula \( x = \frac{mg}{k} \), we find the displacement when the forces are in equilibrium. For maximum compression, we equate the initial kinetic energy to the potential energy stored in the spring: \( \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \). Solving these gives us two different values for spring displacement.
energy conversion in springs
Springs are excellent examples of energy conversion systems. They demonstrate the transformation between kinetic energy and potential energy. When a spring compresses, kinetic energy from a moving object is transferred into the spring, storing potential energy. When released, this potential energy is transformed back into kinetic energy, often propelling the object in motion. This conversion process highlights the law of conservation of energy, as seen in the exercise where the box's kinetic energy is entirely converted into the potential energy of the compressed spring. Understanding this conversion process is crucial in mechanics, especially when solving problems involving oscillatory motion or systems with elastic components.

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Most popular questions from this chapter

Two physical pendulums (not simple pendulums) are made from meter sticks that are suspended from the ceiling at one end. The sticks are uniform and are identical in all respects, except that one is made of wood (mass \(=0.17 \mathrm{kg}\) ) and the other of metal (mass \(=0.85 \mathrm{kg}\) ). They are set into oscillation and execute simple harmonic motion. Determine the period of (a) the wood pendulum and (b) the metal pendulum.

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 0.60-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 5.70 to 4.80 m/s. What is the spring constant of the spring?

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 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 \(f=3.0 \mathrm{Hz} ?\)

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