/*! 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 79 Consider a block of mass 0.200 k... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider a block of mass 0.200 kg attached to a spring of spring constant \(100 \mathrm{N} / \mathrm{m}\). The block is placed on a frictionless table, and the other end of the spring is attached to the wall so that the spring is level with the table. The block is then pushed in so that the spring is compressed by \(10.0 \mathrm{cm} .\) Find the speed of the block as it crosses (a) the point when the spring is not stretched, (b) \(5.00 \mathrm{cm}\) to the left of point in (a), and (c) \(5.00 \mathrm{cm}\) to the right of point in (a).

Short Answer

Expert verified
The speed of the block at different points are: (a) 1.00 m/s when the spring is not stretched, (b) 0.866 m/s when the spring is stretched 5.00 cm to the left of point (a), and (c) 0.866 m/s when the spring is stretched 5.00 cm to the right of point (a).

Step by step solution

01

Write down the given information and the law of conservation of energy

Mass of the block: \(m = 0.200 \mathrm{kg}\) Spring constant: \(k = 100 \mathrm{N/m}\) Initial compression of the spring: \(x_0 = 10.0 \mathrm{cm} = 0.100 \mathrm{m}\) The law of conservation of energy states that the total mechanical energy (kinetic energy + potential energy) remains constant in the absence of external forces.
02

Set up the equation for the conservation of energy

At the initial state (spring compressed), all energy is stored as potential energy in the spring: \(E_{initial} = \frac{1}{2} kx_0^2\) At any later point, the energy will be in the form of kinetic energy of the block and potential energy in the spring. Let the compression at some point be \(x\), so the potential energy is \(\frac{1}{2} kx^2\) and the kinetic energy is \(\frac{1}{2} mv^2\). Therefore, \(E_{total} = \frac{1}{2} kx^2 + \frac{1}{2} mv^2\) According to the law of conservation of energy, \(E_{initial} = E_{total}\)
03

Calculate speed at point (a) – spring not stretched

At point (a), the spring is not stretched, so \(x = 0\). The equation for conservation of energy becomes: \(\frac{1}{2} kx_0^2 = \frac{1}{2} mv_a^2\) Solve for \(v_a\): \(v_a = \sqrt{\frac{kx_0^2}{m}}\) Plug in the given values: \(v_a = \sqrt{\frac{100(0.100)^2}{0.200}}\) \(v_a = 1.00 \, \mathrm{m/s}\) The speed of the block at point (a) is 1.00 m/s.
04

Calculate speed at point (b) – 5.00 cm to the left of point a

At point (b), the spring is stretched 5 cm to the left, so \(x = -0.050 \, \mathrm{m}\). Applying the energy conservation equation: \(\frac{1}{2} kx_0^2 = \frac{1}{2} kx^2 + \frac{1}{2} mv_b^2\) Solve for \(v_b\): \(v_b = \sqrt{\frac{k(x_0^2 - x^2)}{m}}\) Plug in the given values: \(v_b = \sqrt{\frac{100((0.100)^2 - (-0.050)^2)}{0.200}}\) \(v_b = 0.866 \, \mathrm{m/s}\) The speed of the block at point (b) is 0.866 m/s.
05

Calculate speed at point (c) – 5.00 cm to the right of point a

At point (c), the spring is again stretched by 5 cm but to the right, so \(x = 0.050 \, \mathrm{m}\). Applying the energy conservation equation: \(\frac{1}{2} kx_0^2 = \frac{1}{2} kx^2 + \frac{1}{2} mv_c^2\) Solve for \(v_c\): \(v_c = \sqrt{\frac{k(x_0^2 - x^2)}{m}}\) Plug in the given values: \(v_c = \sqrt{\frac{100((0.100)^2 - (0.050)^2)}{0.200}}\) \(v_c = 0.866 \, \mathrm{m/s}\) The speed of the block at point (c) is 0.866 m/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hooke's Law
When studying spring systems, one fundamental principle is Hooke's law. It states that the force required to compress or extend a spring by a certain distance is proportional to that distance. The formula is expressed as

\[ F = -kx \]

where \( F \) is the force exerted by the spring, \( k \) is the spring constant, a measure of the spring's stiffness, and \( x \) is the displacement from the spring's equilibrium position. This law applies as long as the spring is not stretched beyond its elastic limit, meaning the spring can return to its initial shape after being deformed. In our textbook exercise, we used Hooke's law to relate the spring constant and the compression of the spring to the force on the block.
Elastic Potential Energy
Elastic potential energy is the form of potential energy that is stored in elastic materials when they are stretched or compressed. It's significant in the study of conservation of energy in spring systems. The formula to calculate the elastic potential energy in a spring is

\[ E_{\text{elastic}} = \frac{1}{2} kx^2 \]

The term \( \frac{1}{2} \) appears because the energy is proportional to the square of the displacement, and \( k \) represents the spring constant. \( x \) stands for the distance the spring has been deformed from its equilibrium position. In the exercise's context, we calculated the initial elastic potential energy of the spring when it was compressed, and we also considered changes to this energy as the spring extended or compressed more at different points.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It is calculated by the equation

\[ E_{\text{kinetic}} = \frac{1}{2} mv^2 \]

with \( m \) being the mass of the object and \( v \) its velocity. For our block attached to the spring, we analyzed how the kinetic energy varied as the spring went from being compressed to uncompressed. At any point in time, the sum of the block's kinetic energy and the spring's elastic potential energy must be constant, assuming no external forces like friction are at work. This concept demonstrates the conservation of energy, which is pivotal in understanding the problem at hand.
Spring Constant
The spring constant, denoted by the symbol \( k \), quantifies the stiffness of a spring. It can be found from Hooke's law and is measured in Newtons per meter (N/m). The stiffer the spring, the greater its spring constant. In our exercise, we were given a spring constant of \( 100 \mathrm{N/m} \), which was used to differentiate the force applied by the spring at various levels of compression or extension. The spring constant's role in the equations of elastic potential energy and Hooke's law helped us determine the relationship between the displacement of the spring and the energies involved in the system.
Harmonic Motion
Harmonic motion, or simple harmonic motion (SHM), refers to repetitive oscillation back and forth through an equilibrium position. A system where a mass is attached to a spring exhibits SHM if the spring follows Hooke’s law and there are no external forces like friction affecting the motion. Such a system's motion can be described by sinusoidal functions of time.

In the context of the exercise, when the block is released from a compressed state, it oscillates around the equilibrium position where the spring is at its natural length. This motion includes cycles of transferring energy back and forth between kinetic and elastic potential forms, perfectly illustrating the concept of conservation of energy in physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A block of mass 300 g is attached to a spring of spring constant \(100 \mathrm{N} / \mathrm{m}\). The other end of the spring is attached to a support while the block rests on a smooth horizontal table and can slide freely without any friction. The block is pushed horizontally till the spring compresses by \(12 \mathrm{cm}\) and then the block is released from rest. (a) How much potential energy was stored in the block-spring support system when the block was just released? (b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched. (c) Determine the speed of the block when it has traveled a distance of 20 \(\mathrm{cm}\) from where it was released.

Shown below is a box of mass \(m_{1}\) that sits on a frictionless incline at an angle above the horizontal \(\theta=30^{\circ} .\) This box is connected by a relatively massless string, over a frictionless pulley, and finally connected to a box at rest over the ledge, labeled \(m_{2} .\) If \(m_{1}\) and \(m_{2}\) are a height \(h\) above the ground and \(m_{2}>>m_{1}:\) (a) What is the initial gravitational potential energy of the system? (b) What is the final kinetic energy of the system?

\(\begin{array}{lllll}\text { 52. } & \text { In } & \text { the } & \text { cartoon } & \text { movie } & \text { Pocahontas }\end{array}\) (https://openstaxcollege.org/l/21pocahontclip) Pocahontas runs to the edge of a cliff and jumps off, showcasing the fun side of her personality. (a) If she is running at \(3.0 \mathrm{m} / \mathrm{s}\) before jumping off the cliff and she hits the water at the bottom of the cliff at \(20.0 \mathrm{m} / \mathrm{s}\), how high is the cliff? Assume negligible air drag in this cartoon. (b) If she jumped off the same cliff from a standstill, how fast would she be falling right before she hit the water?

A single force \(F(x)=-4.0 x\) (in newtons) acts on a 1.0-kg body. When \(x=3.5 \mathrm{m},\) the speed of the body is 4.0 \(\mathrm{m} / \mathrm{s} .\) What is its speed at \(x=2.0 \mathrm{m} ?\)

Two people observe a leaf falling from a tree. One person is standing on a ladder and the other is on the ground. If each person were to compare the energy of the leaf observed, would each person find the following to be the same or different for the leaf, from the point where it falls off the tree to when it hits the ground: (a) the kinetic energy of the leaf; (b) the change in gravitational potential energy; (c) the final gravitational potential energy?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.