/*! 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 84 A certain spring is found \(n o ... [FREE SOLUTION] | 91Ó°ÊÓ

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A certain spring is found \(n o t\) to conform to Hooke's law. The force (in newtons) it exerts when stretched a distance \(x\) (in meters) is found to have magnitude \(52.8 x+38.4 x^{2}\) in the direction opposing the stretch. (a) Compute the work required to stretch the spring from \(x=0.500 \mathrm{~m}\) to \(x=1.00 \mathrm{~m} .\) (b) With one end of the spring fixed, a particle of mass \(2.17 \mathrm{~kg}\) is attached to the other end of the spring when it is stretched by an amount \(x=1.00 \mathrm{~m}\). If the particle is then released from rest, what is its speed at the instant the stretch in the spring is \(x=0.500 \mathrm{~m} ?(\mathrm{c})\) Is the force exerted by the spring conservative or nonconservative? Explain.

Short Answer

Expert verified
(a) 31.0 J; (b) 5.34 m/s; (c) The force is conservative.

Step by step solution

01

Understand the Problem and Extract Useful Information

We have a spring that does not follow Hooke's Law. Instead, the force exerted by the spring is given by the expression \( F(x) = 52.8x + 38.4x^2 \). We need to calculate the work done when stretching the spring from \( x = 0.5 \) to \( x = 1.0 \) meters, find the speed of a mass attached to the spring as it moves from \( x = 1.0 \) to \( x = 0.5 \) meters, and determine if the spring force is conservative.
02

Calculate the Work Done Using Integration

The work done by a force, particularly a variable force like this, is obtained by integrating the force over the distance it acts. For part (a), we compute the work \( W \) with\[ W = \int_{0.5}^{1.0} (52.8x + 38.4x^2) \, dx. \]First, integrate:\[ \int (52.8x + 38.4x^2) \, dx = 26.4x^2 + 12.8x^3 + C. \]Now, evaluate from 0.5 to 1.0:\[ W = \left[ 26.4(1)^2 + 12.8(1)^3 \right] - \left[ 26.4(0.5)^2 + 12.8(0.5)^3 \right] \]\[ W = (26.4 + 12.8) - \left(26.4 \times 0.25 + 12.8 \times 0.125\right) \]\[ W = 39.2 - (6.6 + 1.6) = 39.2 - 8.2 = 31.0 \, \text{Joules}. \]
03

Apply Energy Conservation for Calculating Speed

When the spring is stretched from \( x = 1.0 \) m to \( x = 0.5 \) m, and a particle is attached, we use the work-energy theorem. The change in kinetic energy is equal to the negative of the work done by the spring, because it's in the opposite direction.The initial kinetic energy (KE) at \( x = 1.0 \) is 0 (since it starts at rest), and we calculate the change in potential energy (work) from \( x = 1.0 \) to \( x = 0.5 \):\[ \Delta KE = W = -\int_{0.5}^{1.0} (52.8x + 38.4x^2) \, dx = -31.0 \, \text{J}, \]Now, let the final kinetic energy (at \( x = 0.5 \) m) be \( \frac{1}{2}mv^2 = 31.0 \). Solve for \( v \):\[ \frac{1}{2} \times 2.17 \times v^2 = 31.0 \]\[ v^2 = \frac{62.0}{2.17} \]\[ v = \sqrt{28.57} \approx 5.34 \, \text{m/s}. \]
04

Determine if the Force is Conservative

For a force to be conservative, the work done must depend only on the initial and final positions, not on the path taken. Since the force here derives from a potential energy function (as shown by integrating the force to find work done), it is indeed conservative.

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Key Concepts

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

Work and Energy
Understanding the concept of work and energy is essential when dealing with non-Hookean springs. **Work** in physics refers to the energy transferred by a force acting over a distance. In the context of our exercise, the work done is when the spring's force is applied across the movement span from 0.5 meters to 1.0 meters.
The force in this case is variable, changing as the spring stretches, and is described mathematically by the function: \[ F(x) = 52.8x + 38.4x^2 \]To calculate the work done by this non-constant force, we integrate it over the distance. This means we find the area under the curve of the force function over the specified range. The integration process allows us to compute the work required to stretch the spring precisely: \[ W = \int_{0.5}^{1.0} (52.8x + 38.4x^2) \, dx = 31.0 \, \text{Joules}. \]So, when students perform this integration, they quantify the energy that was necessary to stretch the spring by that given extent.
Conservative Forces
Conservative forces are forces where the work done depends solely on the starting and the ending points, not on the path taken. This is a fundamental aspect of physics, making them predictable and reversible. Examples of conservative forces include gravity and elastic spring force.
In our exercise, we need to determine whether the force exerted by this non-Hookean spring is conservative. It is key to note that for a force to be conservative, it needs to be derivable from a potential energy function.
During the step-by-step solution, by integrating the force function, we notice that the work calculation depends only on the initial and final stretching points, not any path-dependent characteristics. Therefore, because the force works in such a way that can be described by the potential energy over displacement, it indeed qualifies as conservative. Understanding this helps streamline problem-solving by allowing energy conservation methods to be applied effectively in these scenarios.
Kinetic Energy
Kinetic Energy (KE) is the energy an object possesses due to its motion. It is given by the formula:\[ KE = \frac{1}{2} m v^2 \] where \( m \) is mass and \( v \) is velocity. In the exercise, after determining the work done by the spring as it moves between two points, we utilize the conservation of mechanical energy.
Initially, the mass at the spring's end is at rest with zero kinetic energy. When released and the spring contracts, this stored potential energy gets converted into kinetic energy. The change in kinetic energy amounts to the negative of the work done by the spring on the mass because it opposes the movement.
By substituting into our kinetic energy equation, we can find the velocity of the mass:\[ \frac{1}{2} \times 2.17 \, \mathrm{kg} \times v^2 = 31.0 \, \mathrm{J} \]This results in a speed \( v \approx 5.34 \, \mathrm{m/s} \).
This transformation from potential to kinetic exemplifies energy conservation principles and highlights the practical application of work and energy concepts in real-world physics problems.

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