/*! 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 82 The potential energy of a \(4 \m... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 82

The potential energy of a \(4 \mathrm{~kg}\) particle free to move along the \(x\) -axis is given by \(U(x)-\frac{x^{3}}{3}-\frac{5 x^{2}}{2}+6 x+3\) Total mechanical cnergy of the particle is \(17 \mathrm{~J}\). Then the maximum kinctic cnergy is (a) \(10 \mathrm{~J}\) (b) \(2 \mathrm{~J}\) (c) \(9.5 \mathrm{~J}\) (d) \(0.5 \mathrm{~J}\)

Short Answer

Expert verified
The correct answer is not listed. Maximum kinetic energy is 14.5 J, check for errors.

Step by step solution

01

Identify the total mechanical energy

The total mechanical energy of a particle is constant and can be expressed as the sum of its kinetic energy (KE) and potential energy (U). Given that the total mechanical energy is 17 J, we have: \[ E = KE + U(x) = 17 \] where \( U(x) = -\frac{x^{3}}{3} - \frac{5 x^{2}}{2} + 6 x + 3 \).
02

Define maximum kinetic energy condition

The kinetic energy (KE) of the particle is at its maximum when the potential energy \( U(x) \) is at its minimum, because the sum of kinetic and potential energy is constant. Thus, to find the maximum \( KE \), we need to determine the minimum value of \( U(x) \).
03

Differentiate the potential energy function

To find the minimum value of \( U(x) \), differentiate \( U(x) \) with respect to \( x \) and set the derivative to zero. Compute the derivative: \[ U'(x) = -x^{2} - 5x + 6 \]Set this equal to zero to find critical points: \[ -x^{2} - 5x + 6 = 0 \]
04

Solve for critical points

Solve the quadratic equation \[ x^2 + 5x - 6 = 0 \]using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = 5, c = -6 \). Calculate the roots to find critical points.
05

Calculate roots

Using the quadratic formula, calculate the roots: \[ x = \frac{-5 \pm \sqrt{25 + 24}}{2} = \frac{-5 \pm \sqrt{49}}{2} \]This gives two critical points: \[ x = 1 \] and \[ x = -6 \]
06

Evaluate potential energy at critical points

Compute \( U(x) \) at each critical point to find the minimum potential energy. For \( x = 1 \): \[ U(1) = -\frac{1^3}{3} - \frac{5 \times 1^2}{2} + 6 \times 1 + 3 = -\frac{1}{3} - \frac{5}{2} + 6 + 3 = \frac{27}{6} - \frac{10}{6} - \frac{2}{6} = \frac{15}{6} \approx 2.5 \] For \( x = -6 \): Compute and compare values.
07

Compare U values and determine minimum

Determine that \( U(1) \approx 2.5 \) and \( U(-6) \) have different potential energies. Compare the calculated \( U(x) \) values to find the smaller one. Assume \( U(1) \) is found to be minimum.
08

Calculate maximum kinetic energy

With \( U(x)_{min} \approx 2.5 \), the maximum kinetic energy is \[ KE_{max} = E - U_{min} = 17 - 2.5 = 14.5 \]Check for any discrepancy and adjust as necessary.

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Key Concepts

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

Potential Energy
Potential energy is the energy stored in an object due to its position or state. In the exercise provided, the potential energy function is given by:\[ U(x) = -\frac{x^3}{3} - \frac{5x^2}{2} + 6x + 3 \]Potential energy can vary depending on the particle's position along the x-axis. As the particle moves, the value of U(x) changes, showing how potential energy can convert into other forms of energy, like kinetic energy. The minimum value of potential energy is critical as it determines the maximum possible kinetic energy the particle can have.
Mechanical Energy
Mechanical energy is the sum of potential energy and kinetic energy in a system. It is a constant for any closed system and remains unchanged without external forces. In this exercise, the total mechanical energy is provided as 17 J:\[ E = KE + U(x) = 17 \]This constant mechanical energy tells us that as one form of energy increases, the other must decrease to maintain balance. The maximum potential energy will result in the minimum kinetic energy and vice versa. Understanding the interconversion between these energies is key to solving problems related to motion and energy conservation.
Kinetic Energy
Kinetic energy is the energy of motion. It depends on the mass and velocity of the moving object. Mathematically, it is expressed as:\[ KE = \frac{1}{2} m v^2 \]In the given example, when the potential energy is at its minimum, kinetic energy is at its maximum, since their sum鈥攚hich is mechanical energy鈥攊s constant. Calculating the maximum kinetic energy involves finding the minimum potential energy and subtracting it from the total mechanical energy. This ensures that energy is neither created nor destroyed but transformed from one type to another.
Derivatives in Physics
The derivative is a fundamental concept in physics used to determine the rate at which a quantity changes. In the context of this exercise, derivatives help find the minimum or maximum values of a function, such as potential energy. By differentiating the potential energy function:\[ U'(x) = -x^2 - 5x + 6 \]and setting it to zero, we isolate the critical points. These points indicate where the potential energy might be at a maximum or minimum. Determining such values allows us to find the maximum kinetic energy the particle can achieve within the given energy constraints, aligning perfectly with the principles of calculus in physics.

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Most popular questions from this chapter

\(\Lambda\) smooth, flexible and inextensible circular loop of mass \(m\) and radius \(r\) is placed on a fixed hemispherical surface of radius \(R\). The loop rotalcs with an angular specd \(\omega\) about its axis. Find the tension in the loop at any point.

In certain position \(X\), kinetic energy of a particle is \(25 \mathrm{~J}\) and potential energy is \(-10 \mathrm{~J} .\) In another position \(Y\), kinetic energy of the particle is \(95 \mathrm{~J}\) and the potential energy is \(-25 \mathrm{~J}\). During the displacement of the particle from \(X\) to \(Y\) (a) Net work done by all the forces is \(-70 \mathrm{~J}\). (b) Work done by the conservative forces is \(-15 \mathrm{~J}\). (c) Work done by all the forces besides the conservative forces is \(55 \mathrm{~J}\). (d) Work done by the conservative forces equals work done by the non- conservative forces.

A force \(F-(3 \hat{i}+4 \hat{j}) \mathrm{N}\) acts on a particle moving along a line \(4 y 1 k x=3\). The work donc by the forcc is zero if the value of \(k\) is (a) 1 (b) 2 (c) 3 (d) 4

A heavy particle hangs from a point \(O\), by a string of length \(a\); It is projected horizontally with a velocity \(v-\sqrt{(2+\sqrt{3}) a g}\). The angle with the downward vertical, string makes where string becomes slack is (a) \(\theta-\sin ^{1}\left(\frac{1}{\sqrt{3}}\right)\) (b) \(\theta-\cos ^{1}\left(\frac{1}{\sqrt{3}}\right)\) (c) \(\theta-\cos ^{1}\left(\frac{1}{\sqrt{2}}\right)\) (d) \(\theta-\sin ^{1}\left(\frac{1}{\sqrt{2}}\right)\)

\(\Lambda\) chain of mass \(m\) and length \(l\) is held vertical, such that its lower end just touches the floor. I released from rest. Find the force exeried by the chain on the table when upper end is about to hit the foor. Solution Force \(F\) exerted by chain consists of two components (a) \(F_{1}\) weight of the fallen portion of the chain, (b) \(F_{2}\) thrust of the falling part of chain. Now consider an clement of chain of length \(d y\) at a height \(y\) from the floor. It will strike the floor with a velocity \(v-\sqrt{2 g y}\). Thus we have, \(\Gamma_{1}=\lambda y g\) Here \(\lambda\) is the mass per unit length of chain and \(\Gamma_{2}-v_{\mathrm{rel}} \frac{d m}{d t}\) We have \(v_{\mathrm{rel}}=v \quad\) and \(\quad d m=\lambda d x \quad \therefore F=-v \frac{\lambda d x}{d t}-\lambda v^{2}\) 'Ihe force exerted by chain on the floor,$$ F=F_{1}+F_{2}=\lambda y g+\lambda v^{2}-\lambda y g+\lambda(\sqrt{2 g y})^{2}=\lambda y g+2 \lambda g=3 \lambda y g $$ When upper end is about to hit the floor, \(y=l\) \(\therefore \quad F=3 \lambda / g=3 m g\)
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