/*! 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 27 A spring is hung from the ceilin... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
(a) Spring constant is \( 58.8 \, \text{N/m} \), (b) angular frequency is \( 11.42 \, \text{rad/s} \).

Step by step solution

01

Understanding the Problem

The block stretches the spring until it momentarily stops, reaching maximum displacement. This problem can be solved using Hooke's Law and the concept of potential energy changes in the spring.
02

Identify Given Values

The mass of the block is given as \( m = 0.450 \, \text{kg} \), and the displacement of the block from equilibrium is \( ext{x} = 0.150 \, \text{m} \). We'll use these values to find the spring constant and angular frequency.
03

Use Energy Conservation to Find Spring Constant

The potential energy stored in the spring at maximum displacement equals the gravitational potential energy lost by the block. Use the equation: \( mgx = \frac{1}{2}kx^2 \) to find the spring constant \( k \). Simplifying gives \( k = \frac{2mg}{x} \).
04

Calculate the Spring Constant

Substituting values into the equation \( k = \frac{2mg}{x} \), we have \( k = \frac{2 \times 0.450 \, \text{kg} \times 9.8 \, \text{m/s}^2}{0.150 \, \text{m}} = 58.8 \, \text{N/m} \).
05

Calculate Angular Frequency

Use the formula for angular frequency \( \omega = \sqrt{\frac{k}{m}} \). Substituting the values, \( \omega = \sqrt{\frac{58.8 \, \text{N/m}}{0.450 \, \text{kg}}} \approx 11.42 \, \text{rad/s} \).
06

Conclusion

Thus, the spring constant of the spring is \( 58.8 \, \text{N/m} \) and the angular frequency of the block's vibrations is approximately \( 11.42 \, \text{rad/s} \).

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Key Concepts

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

Hooke's Law
Hooke's Law is a fundamental principle for understanding how springs behave under force. Imagine a spring like a rubber band; it can stretch and compress. This law states that the force needed to stretch or compress a spring by a distance is directly proportional to that distance. It can be mathematically represented as \[ F = kx \]where:
  • \( F \) is the force applied to the spring,
  • \( k \) is the spring constant, which tells us how stiff the spring is, and
  • \( x \) is the displacement or change in length of the spring.
The spring constant \( k \) is a measure of the force required to stretch or compress the spring by one unit of distance. In other words, a bigger \( k \) means a stiffer spring. For the given exercise, knowing the spring stretches to a certain length gives us a chance to calculate this constant, revealing the spring's stiffness. Understanding Hooke’s Law is vital as it lays the groundwork for further calculations involving the spring's potential energy when it stretches.
Energy Conservation
Energy conservation is a key concept in physics, suggesting that energy cannot be created or destroyed, only transformed from one form to another. In our spring problem, energy conservation is handy to find how much energy shifts between two forms as the block moves.

Initially, the block has potential energy due to its position. As it falls, this potential energy turns into kinetic energy and eventually into elastic potential energy in the spring. Earth's gravitational force (weight of the block) converts to the elastic potential energy stored in the spring. The equation used for this energy transformation is:\[ mgx = \frac{1}{2}kx^2 \]where:
  • \( m \) is the mass of the block
  • \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
  • \( x \) is the displacement
  • \( k \) is the spring constant
This equation helps us find the spring constant \( k \) by showing how gravitational energy translates into energy stored in the spring, when the block momentarily rests at its lowest point.
Angular Frequency
Angular frequency is crucial to understanding how quickly something oscillates. It tells us how fast an object moves through its cycle in oscillatory motions like those of the block on the spring. It's directly tied to the spring's stiffness and the object's mass attached to it.

The formula for angular frequency \( \omega \) is:\[ \omega = \sqrt{\frac{k}{m}} \]where:
  • \( \omega \) is the angular frequency measured in radians per second,
  • \( k \) is the spring constant, and
  • \( m \) is the mass of the oscillating block.
This relationship reveals that a stiffer spring or a lighter mass results in a higher angular frequency, meaning the block vibrates faster. For problems involving frequency and vibrations, grasping this connection helps in predicting how an object behaves in oscillatory motion. The given exercise uses this formula to calculate how rapidly the block moves up and down, showing the practical applications of these theoretical principles.

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Most popular questions from this chapter

A hand exerciser utilizes a coiled spring. A force of \(89.0 \mathrm{N}\) is required to compress the spring by \(0.0191 \mathrm{m} .\) Determine the force needed to compress the spring by \(0.0508 \mathrm{m}\).

A heavy-duty stapling gun uses a 0.140 -kg metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a "ram spring" \((k=32000 \mathrm{N} / \mathrm{m})\). The mass of this spring may be ignored. The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\) when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

A spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is \(0.80 \mathrm{kg}\), and the spring has a spring constant of \(59 \mathrm{N} / \mathrm{m}\). The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74 .\) How far can the spring be stretched from its unstrained position without the box moving when it is released?

A 70.0 -kg circus performer is fired from a cannon that is elevated at an angle of \(40.0^{\circ}\) above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by \(3.00 \mathrm{m}\) from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as the height of the net into which he is shot. He takes 2.14 s to travel the horizontal distance of \(26.8 \mathrm{m}\) between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

A square plate is \(1.0 \times 10^{-2} \mathrm{m}\) thick, measures \(3.0 \times 10^{-2} \mathrm{m}\) on a side, and has a mass of \(7.2 \times 10^{-2}\) kg. The shear modulus of the material is \(2.0 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) One of the square faces rests on a flat horizontal surface, and the coefficient of static friction between the plate and the surface is 0.91 . A force is applied to the top of the plate, as in Figure \(10.29 a .\) Determine (a) the maximum possible amount of shear stress, (b) the maximum possible amount of shear strain, and (c) the maximum possible amount of shear deformation \(\Delta X\) (see Figure \(10.29 b\) ) that can be created by the applied force just before the plate begins to move.

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