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A \(0.40-\mathrm{kg}\). mass is attached to a spring with a force constant of \(26 \mathrm{N} / \mathrm{m}\) and released from rest a distance of \(3.2 \mathrm{cm}\) from the equilibrium position of the spring. (a) Give a strategy that allows you to find the speed of the mass when it is halfway to the equilibrium position. (b) Use your strategy to find this speed.

Short Answer

Expert verified
The speed is approximately 0.223 m/s.

Step by step solution

01

Understanding the Problem

We need to determine the speed of a mass attached to a spring as it moves towards equilibrium. The mass is initially released from rest, 3.2 cm from equilibrium. We have to find its speed when it's halfway to the equilibrium position.
02

Convert Units to SI

To work with standard units, convert the initial displacement from cm to meters: 3.2 cm = 0.032 m.
03

Set Up Conservation of Energy Equation

The total mechanical energy of the system is conserved. Initially, all energy is potential energy: \[ E_{initial} = \frac{1}{2} k x^2 \] where \( k = 26 \, \text{N/m} \) and \( x = 0.032 \, \text{m} \).
04

Calculate Initial Potential Energy

Calculate the initial potential energy when the mass is displaced by 0.032 m:\[ E_{initial} = \frac{1}{2} \times 26 \, \text{N/m} \times (0.032 \, \text{m})^2 = 0.013312 \, \text{J} \]
05

Determine Position at Halfway to Equilibrium

When the mass is halfway to the equilibrium position, the displacement is half of 0.032 m, so 0.016 m.
06

Calculate Potential Energy at Halfway

Find the potential energy at this halfway point:\[ E_{potential} = \frac{1}{2} \times 26 \, \text{N/m} \times (0.016 \, \text{m})^2 = 0.003328 \, \text{J} \]
07

Use Conservation of Energy to Find Kinetic Energy

Use conservation of energy: initial energy equals the sum of potential and kinetic energy at halfway. Solve for kinetic energy:\[ 0.013312 \, \text{J} = 0.003328 \, \text{J} + E_{kinetic} \] \[ E_{kinetic} = 0.009984 \, \text{J} \]
08

Solve for Velocity Using Kinetic Energy Formula

Kinetic energy is given by \( E_{kinetic} = \frac{1}{2} mv^2 \). Solve for \( v \):\[ v = \sqrt{\frac{2 \, E_{kinetic}}{m}} = \sqrt{\frac{2 \times 0.009984 \, \text{J}}{0.40 \, \text{kg}}} \]\[ v \approx 0.223 \, \text{m/s} \]
09

Conclusion

The speed of the mass when it is halfway to the equilibrium position is approximately 0.223 m/s.

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Key Concepts

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

Conservation of Energy
In simple harmonic motion, the principle of conservation of energy plays a crucial role. It states that the total energy within a closed system remains constant. This means that energy can transform between different types, such as potential energy and kinetic energy, but the sum of these energies stays the same.

In our spring-mass system, when the mass is fully displaced from the equilibrium position, all the energy is stored as potential energy. As the mass moves towards equilibrium, this potential energy transforms into kinetic energy, increasing the speed of the mass. Halfway to the equilibrium, the energy is a combination of both potential and kinetic. This transformation of energy adheres to the conservation of energy principle, ensuring that no energy is lost in the process, assuming it's an ideal system without friction or air resistance.
Spring Constant
The spring constant, often denoted as \( k \), is a measure of a spring's stiffness. It's a key factor in Hooke's Law, which is expressed as \( F = kx \), where \( F \) is the force applied to the spring and \( x \) is the displacement from equilibrium.

A higher spring constant means a stiffer spring that requires more force to compress or stretch. In our problem, the spring constant is given as \( 26 \, \text{N/m} \). This value tells us how responsive the spring is to displacement. In the context of energy, the spring constant affects how potential energy is stored in the system. A greater spring constant leads to more potential energy for a given displacement, which then influences how much kinetic energy is available as the mass moves toward equilibrium.
Kinetic Energy
Kinetic energy is the energy associated with the motion of an object. In our spring system, as the mass moves from rest towards equilibrium, it gains speed and thereby kinetic energy.

The relationship between mass and kinetic energy is described by the formula \( E_{kinetic} = \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity. Halfway to equilibrium, the kinetic energy can be calculated using the conservation of energy principles, subtracting the new potential energy from the total starting energy.

In this exercise, we determined the kinetic energy halfway to the equilibrium point and used it to calculate the mass’s velocity. Understanding kinetic energy helps explain how fast the mass moves and how energy shifts from one form to another in harmonic motion.
Potential Energy
Potential energy in a spring system is the energy stored due to the spring's displacement. It's calculated using the formula \( E_{potential} = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement.

Initially, when the mass is pulled to its maximum displacement, all the system's energy is potential. As the mass begins moving towards the equilibrium position, this energy starts to convert into kinetic energy. At halfway to equilibrium, the potential energy decreases because the displacement is smaller, specifically half of the initial distance in our case.

Understanding potential energy in this context helps explain how the energy distribution changes as the spring returns to its original form, enabling us to find the kinetic energy and thus the speed of the mass during its motion.

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Most popular questions from this chapter

A 0.321 -kg mass is attached to a spring with a force constant of \(13.3 \mathrm{N} / \mathrm{m}\). If the mass is displaced \(0.256 \mathrm{m}\) from equilibrium and released, what is its speed when it is \(0.128 \mathrm{m}\) from equilibrium?

A mass oscillates on a spring with a period \(T\) and an amplitude \(0.48 \mathrm{cm} .\) The mass is at the equilibrium position \(x=0\) at \(t=0,\) and is moving in the positive direction. Where is the mass at the times (a) \(t=T / 8\) (b) \(t=T / 4\) (c) \(t=T / 2\) and (d) \(t=3 T / 4 ?\) (e) Plot your results for parts (a) through (d) with the vertical axis representing position and the horizontal axis representing time.

A mass oscillates on a spring with a period of \(0.73 \mathrm{s}\) and an amplitude of \(5.4 \mathrm{cm}\). Write an equation giving \(x\) as a function of time, assuming the mass starts at \(x=A\) at time \(t=0\). .

A \(0.45-k g\) crow lands on a slender branch and bobs up and down with a period of 1.5 s. An eagle flies up to the same branch, scaring the crow away, and lands. The eagle now bobs up and down with a period of \(4.8 \mathrm{s}\). Treating the branch as an ideal spring, find (a) the effective force constant of the branch and (b) the mass of the eagle.

The acceleration of a block attached to a spring is given by \(a=-\left(0.302 \mathrm{m} / \mathrm{s}^{2}\right) \cos ([2.41 \mathrm{rad} / \mathrm{s}] \mathrm{t})\) (a) What is the frequency of the block's motion? (b) What is the maximum speed of the block? (c) What is the amplitude of the block's motion?

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