91Ó°ÊÓ

Nonconforming Spring 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 \((52.8 \mathrm{~N} / \mathrm{m}) x+\left(38.4 \mathrm{~N} / \mathrm{m}^{2}\right) x^{2}\) in the direction opposing the stretch. (a) Compute the work required to stretch the spring from \(x_{1}=0.500\) \(\mathrm{m}\) to \(x_{2}=1.00 \mathrm{~m} .\) (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 extended by an amount \(x_{2}=1.00 \mathrm{~m}\). If the particle is then released from rest, what is its speed at the instant the spring has returned to the configuration in which the extension is \(x_{1}=0.500 \mathrm{~m} ?(\mathrm{c})\) Is the force exerted by the spring conservative or nonconservative? Explain.

Step by step solution

01

Identify the Force Equation

The force exerted by the spring is given by the equation: \[ F(x) = -\left[(52.8 \, \mathrm{N/m}) x + (38.4 \, \mathrm{N/m^2}) x^2\right] \] The negative sign indicates the force opposes the direction of stretch.

02

Compute the Work (Part a)

Work to stretch the spring from \(x_1 = 0.500 \, \mathrm{m}\) to \(x_2 = 1.00 \, \mathrm{m}\) is given by: \[ W = \int_{x_1}^{x_2} F(x) \, dx \] Substitute \(F(x)\) and integrate: \[ W = -\int_{0.5}^{1.0} \left[(52.8 x + 38.4 x^2)\right] \, dx \] First, integrate the linear term: \[ \int 52.8 x \, dx = 26.4 x^2 \] Next, integrate the quadratic term: \[ \int 38.4 x^2 \, dx = 12.8 x^3 \]

03

Evaluate the Indefinite Integrals

Evaluate the indefinite integrals: \[ W = - \left[ 26.4 x^2 + 12.8 x^3 \right]_{0.5}^{1.0} \] Computing the definite integral involves evaluating these integrals at the boundaries: \[ \left[ 26.4 x^2 + 12.8 x^3 \right]_{0.5}^{1.0} = \left(26.4(1.0)^2 + 12.8(1.0)^3 \right) - \left(26.4(0.5)^2 + 12.8(0.5)^3 \right) \]

04

Calculate Work Done (Part a)

Perform the calculations: \[ 26.4(1.0) + 12.8(1.0) = 39.2 \] \[ 26.4(0.25) + 12.8(0.125) = 6.6 + 1.6 = 8.2 \] Work done, \(W\), is: \[ W = - (39.2 - 8.2) = -31.0 \, \mathrm{J} \]

05

Compute Speed (Part b)

Apply conservation of energy. Initial potential energy stored in the spring when stretched to \(x = 1.0 \mathrm{m}\) converts to kinetic and potential energy when the extension is \(x = 0.5 \mathrm{m}\). Initial potential energy: \[ U_i = 31.0 \, \mathrm{J} \] At x=0.5 m: Kinetic energy (K) + Potential energy (U_f). Potential energy at 0.5 m: \[ U_f = \frac{1}{2} (-8.2) = -4.1 \, \mathrm{J} \] Energy conservation: \[ K = U_i - U_f = 31.0 - (-4.1) = 35.1 \, \mathrm{J} \] Kinetic energy relation: \[ \frac{1}{2} m v^2 = 35.1 \, \mathrm{J} \] Solve for \(v\): \[ v = \sqrt{ \frac{2 \times 35.1}{2.17} } \approx 5.7 \, \mathrm{m/s} \]

06

Determine if the Force is Conservative (Part c)

For a force to be conservative, the work done must not depend on the path taken but only on the initial and final positions. Here, as the spring's in-built force function depends on distance that ensures the path taken between two points for magnitude presents constant expenditure, the force is conservative.

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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 describes a linear relationship between the force exerted by a spring and the displacement caused by that force. Usually written as \[ F = -k x \],where \(F\) is the force exerted by the spring, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. However, the spring in this exercise does not conform to Hooke's Law because its force depends non-linearly on distance. This means that the force equation here includes a term \(x^2\),making it a variable force rather than the simple linear form expressed by Hooke's Law.

Work Done by a Variable Force

To calculate the work done by a variable force, we use the integral of the force over distance. In our case, the variable force is given as \[ F(x) = - \big[(52.8 \ \mathrm{N/m}) x + (38.4 \ \mathrm{N/m^2}) x^2 \big] \].The work \(W\) done to stretch the spring from \(x_1 = 0.5 \ \mathrm{m}\) to \[x_2 = 1.0 \ \mathrm{m}]\,is found using \[ W = -\big[ \frac{26.4}{2} x^2 + \frac{38.4}{3} x^3 \]_{0.5}^{1.0} \]. This integral gives us the total work required to stretch or compress the spring over that interval.

Conservative Forces

A force is termed as conservative if the work done by the force is path-independent and depends only on the initial and final positions. For example, gravitational force and elastic spring force (for linear springs) are conservative. In this exercise, you determine whether the spring force is conservative by examining if the work done is the same regardless of the path taken between points. The spring in this exercise produces consistent results over any path, indicating that it is a conservative force.

Potential Energy

Potential energy (PE) is the stored energy of an object due to its position or configuration. For a spring, PE is at a maximum when the spring is either compressed or stretched. The energy stored in a spring is usually calculated using \[ U = \frac{1}{2} k x^2 \].However, for a nonconforming spring like ours, the potential energy is part of the more complex work-energy relationship derived through integration. In this exercise, we calculated potential energy at different extensions to find initial and final PE, transforming into kinetic energy as the spring returns to a relaxed state.

Kinetic Energy

Kinetic energy (KE) is the energy of motion, described by the equation \[ K = \frac{1}{2} m v^2 \].In our problem, as the spring contracts, potential energy converts to kinetic energy. Initially, the stored potential energy is 31.0 J at \(x = 1.0 \ \mathrm{m}\). When the spring moves to \(x = 0.5 \ \mathrm{m}\), a portion converts into kinetic energy. Using energy conservation \( U_i = K + U_f \),we find the particle’s speed at the instant the spring is partially relaxed. The kinetic energy \[ K = U_i - U_f \]solves to give the particle a speed of approximately 5.7 m/s.