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Chapter 15: Problem 13

A \(1.00-\mathrm{kg}\) object is attached to a horizontal spring. The spring is initially stretched by \(0.100 \mathrm{m},\) and the object is released from rest there. It proceeds to move without friction. The next time the speed of the object is zero is 0.500 s later. What is the maximum speed of the object?

Short Answer

Expert verified
The maximum speed of the object is \(0.10\, m/s\).

Step by step solution

01

Calculate spring constant

We start by first finding the spring constant k. The potential energy, stored in a spring which is stretched or compressed by x distance is \(E = \frac{1}{2}kx^2\). We can solve for k. The potential energy of the spring is initially transformed into kinetic energy of the object, thus we can state \(E_k = E_p\), leading to \(k = \frac{2E_k}{x^2}\). The maximum kinetic energy can be found using the equation \(E_k = \frac{1}{2}mv^2\).
02

Find maximum kinetic energy

From the kinetic energy formula \(E_k = \frac{1}{2}mv^2\), and the anaylsis in step 1, we know that the initial potential energy in the spring is equal to the maximum kinetic energy. The initial potential energy in the spring can be found using the equation \(E_p=\frac{1}{2}kx^2\). We substitute the known values into the equation, getting \(E_k = \frac{1}{2}kx^2 = \frac{1}{2} \times 1.00\, kg \times (0.100\, m)^2 = 0.005\, J\).
03

Find maximum speed

We now find the maximum speed of the object by rearranging the kinetic energy formula. We solve for maximum speed (v) by manipulation of the formula \(E_k = \frac{1}{2}mv^2\) to \(v = \sqrt{\frac{2E_k}{m}}\), and substituting the found values into the equation: \(v = \sqrt{\frac{2 \times 0.005J}{1.00\, kg}} = 0.10\, m/s\).

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Key Concepts

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

Spring Constant
The spring constant, often denoted as \( k \), is a value that describes a spring's stiffness. It plays a crucial role in simple harmonic motion where it can tell us how much force is needed to stretch or compress the spring by a certain amount. The unit of \( k \) is Newton per meter (N/m). In simple harmonic motion, the spring constant plays into finding energies stored within the spring.
  • The formula to calculate the spring constant is: \( E = \frac{1}{2}kx^2 \), where \( E \) is the energy stored, and \( x \) is the displacement from its equilibrium position.
  • You can rearrange this to find \( k \): \( k = \frac{2E}{x^2} \).
  • This means, the stiffer or tighter the spring is, the larger the \( k \) value.
In the given exercise, the initial potential energy gets transformed to kinetic energy when the spring is stretched.
Potential Energy
Potential energy in simple harmonic motion often refers to the energy stored in a spring when it is displaced from its equilibrium position. This energy depends on the spring constant and the displacement. You can imagine this as the spring "wanting" to return to its normal length.
  • The potential energy stored in a stretched or compressed spring is calculated by \( E_p = \frac{1}{2}kx^2 \).
  • The larger the displacement \( x \), the greater the potential energy stored.
  • This stored energy will convert fully into kinetic energy when the spring returns to its equilibrium.
In the exercise, when we initially stretch the spring 0.100 m, it's storing energy that will later convert into the motion of the object.
Kinetic Energy
Kinetic energy is the energy of motion. When the potential energy in a spring transforms back into kinetic energy, the object it was connected to starts moving. The speed of this object can provide insights into the equality of the energies.
  • The formula for kinetic energy is \( E_k = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.
  • In the context of simple harmonic motion, the kinetic energy reaches its maximum when potential energy is zero, meaning at equilibrium, the object moves fastest.
  • We solved for maximum velocity \( v \) using \( v = \sqrt{\frac{2E_k}{m}} \), where \( E_k \) is the kinetic energy calculated previously.
In the problem at hand, understanding kinetic energy helps us calculate the object's maximum speed as it moves back to its equilibrium position.

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Most popular questions from this chapter

A block-spring system oscillates with an amplitude of \(3.50 \mathrm{cm} .\) If the spring constant is \(250 \mathrm{N} / \mathrm{m}\) and the mass of the block is \(0.500 \mathrm{kg},\) determine (a) the mechanical energy of the system, (b) the maximum speed of the block, and (c) the maximum acceleration.

A weight of \(40.0 \mathrm{N}\) is suspended from a spring that has a force constant of \(200 \mathrm{N} / \mathrm{m} .\) The system is undamped and is subjected to a harmonic driving force of frequency \(10.0 \mathrm{Hz}\) resulting in a forced-motion amplitude of \(2.00 \mathrm{cm} .\) Determine the maximum value of the driving force.

(a) A hanging spring stretches by \(35.0 \mathrm{cm}\) when an object of mass \(450 \mathrm{g}\) is hung on it at rest. In this situation, we define its position as \(x=0 .\) The object is pulled down an additional \(18.0 \mathrm{cm}\) and released from rest to oscillate without friction. What is its position \(x\) at a time 84.4 s later? (b) What If? A hanging spring stretches by \(35.5 \mathrm{cm}\) when an object of mass \(440 \mathrm{g}\) is hung on it at rest. We define this new position as \(x=0 .\) This object is also pulled down an additional \(18.0 \mathrm{cm}\) and released from rest to oscillate without friction. Find its position 84.4 s later. (c) Why are the answers to (a) and (b) different by such a large percentage when the data are so similar? Does this circumstance reveal a fundamental difficulty in calculating the future? (d) Find the distance traveled by the vibrating object in part (a). (e) Find the distance traveled by the object in part (b).

A simple harmonic oscillator takes \(12.0 \mathrm{s}\) to undergo five complete vibrations. Find (a) the period of its motion, (b) the frequency in hertz, and (c) the angular frequency in radians per second.

A \(0.500-\mathrm{kg}\) object attached to a spring with a force constant of \(8.00 \mathrm{N} / \mathrm{m}\) vibrates in simple harmonic motion with an amplitude of \(10.0 \mathrm{cm} .\) Calculate (a) the maximum value of its speed and acceleration, (b) the speed and acceleration when the object is \(6.00 \mathrm{cm}\) from the equilibrium position, and (c) the time interval required for the object to move from \(x=0\) to \(x=8.00 \mathrm{cm}\).
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