/*! 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 1 A single conservative force \(F(... [FREE SOLUTION] | 91Ó°ÊÓ

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A single conservative force \(F(x)\) acts on a \(1.0-\mathrm{kg}\) particle that moves along the \(x\)-axis. The potential energy \(U(x)\) is given by \(U(x)=20+(x-2)^{2}\) where \(x\) is in meters. At \(x=5.0 \mathrm{~m}\), the particle has a kinetic energy of \(20 \mathrm{~J}\). What is the mechanical energy of the system? (1) \(35 \mathrm{~J}\) (2) \(64 \mathrm{~J}\) (3) \(86 \mathrm{~J}\) (4) \(49 \mathrm{~J}\)

Short Answer

Expert verified
The mechanical energy of the system is 49 J.

Step by step solution

01

Identify the Given Information

We are given that the potential energy function is \( U(x) = 20 + (x-2)^2 \). The kinetic energy of the particle at \( x = 5.0 \) m is 20 J. Our task is to calculate the total mechanical energy of the system.
02

Calculate the Potential Energy at x = 5.0 m

To find the potential energy at \( x = 5.0 \) m, use the given potential energy function. Substitute \( x = 5.0 \) into the equation:\[U(5.0) = 20 + (5.0 - 2)^2 = 20 + 3^2 = 20 + 9 = 29 \, \text{J}\]So, the potential energy at \( x = 5.0 \) m is 29 J.
03

Calculate the Total Mechanical Energy

The mechanical energy of the system is the sum of the kinetic and potential energy. We are given the kinetic energy \( K \) is 20 J and the potential energy we just calculated is 29 J.\[E = K + U = 20 \, \text{J} + 29 \, \text{J} = 49 \, \text{J}\]Thus, the total mechanical energy is 49 J.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a crucial part of understanding mechanical energy. It's the energy that an object has due to its motion. When you see something moving, it has kinetic energy because of the speed it possesses. The faster the object goes, the more kinetic energy it holds. This relationship is captured in the formula for kinetic energy, which is given by:\[ K = \frac{1}{2}mv^2 \]where \(m\) is the mass of the object and \(v\) is its velocity.
An important thing to remember here is that kinetic energy depends on both the mass of the object and the square of its velocity. This means that even small changes in speed can significantly impact the kinetic energy of an object! In the original exercise, the particle at \(x = 5.0 \,m\) has a kinetic energy of 20 J. This value is a part of the total mechanical energy of the system, which also includes potential energy.
Potential Energy
Potential energy is the energy stored in an object due to its position or arrangement. It's like a battery that has the potential to do work when needed. In mechanics, potential energy is often related to the forces acting on an object. One of the simplest examples of potential energy is gravitational potential energy, which is based on an object's height. However, in our case with the particle exercise, we are dealing with potential energy along the \(x\)-axis, described by the equation:\[ U(x) = 20 + (x-2)^2 \]This formula illustrates how potential energy changes with the position \(x\) of the particle. At \(x = 5.0 \,m\), substituting into the formula gives us a potential energy of 29 J.
This energy is part of the total mechanical energy of the system, complementing the particle's kinetic energy. It's essential to appreciate how the particle's position influences its potential energy, which, when added to kinetic energy, offers the complete picture of mechanical energy.
Conservative Force
Understanding conservative forces is key to grasping how energy is conserved within a system. A conservative force is one where the work done is independent of the path taken, depending only on the initial and final positions. This means that energy lost in one form can be entirely recovered in another, such as converting potential energy to kinetic energy and vice versa.
The force involved in this exercise is conservative, which allows us to determine mechanical energy simply by adding kinetic and potential energies. Since mechanical energy remains constant in such a system, knowing these energies at one position (like \(x = 5.0 \,m\)) helps us determine them at any other point, provided no non-conservative forces interfere, like friction.
This property simplifies calculations and helps predict motion because the total mechanical energy, composed of kinetic and potential energies, doesn’t change unless external forces come into play.

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Most popular questions from this chapter

Mark the correct statement(s). (1) The work-energy theorem is valid only for particles (2) The work-energy theorem is an invariant law of physics. (3) The work-energy theorem is valid only in inertial frames of reference. (4) The work-energy theorem can be applied in noninertial frames of reference too.

A collar \(B\) of mass \(2 \mathrm{~kg}\) is constrained to move along a horizontal smooth and fixed circular track of radius \(5 \mathrm{~m}\). The spring lying in the plane of the circular track and having spring constant \(200 \mathrm{Nm}^{-1}\) is undeformed when the collar is at \(A\). If the collar starts from rest at \(B\), the normal reaction exerted by the track on the collar when it passes through \(A\) is (1) \(360 \mathrm{~N}\) (2) \(720 \mathrm{~N}\) (3) \(1440 \mathrm{~N}\) (4) \(2880 \mathrm{~N}\)

A system consists of two identical cubes, each of mass \(3 \mathrm{~kg}\), linked together by a compressed weightless spring of force constant \(1000 \mathrm{~N} \mathrm{~m}^{-1}\). The cubes are also connected by a thread which is burnt at a certain moment. At what minimum value of initial compression \(x_{0}\) (in \(\mathrm{cm}\) ) of the spring will the lower cube bounce up after the thread is burnt through?

A body of mass \(M\) was slowly hauled up a rough hill by a force \(F\) which at each point was directed along a tangent to the hill. Work done by the force (1) is independent of the shape of trajectory (2) depends upon the vertical component of displacemert but is independent of horizontal component (3) depends upon both the components (4) does not depend upón the coefficient of friction

A chain of length \(l\) and mass \(m\) lies on the surface of a smooth sphere of radius \(R>l\) with one end tied to the top of the sphere. $$ \begin{array}{|c|c|} \hline {\begin{array}{c} \text { Column I } \\ \end{array}} & {\begin{array}{c} \text { Column II } \\ \end{array}} \\ \hline \text { i. } \text {Gravitational potential energy w.r.t. centre of the sphere} & \text { a. } \text{\(\frac{\operatorname{Rg}}{l}\left[1-\cos \left(\frac{l}{R}\right)\right]\)} \\ \hline \text { ii. } \text{The chain is released and slides down, its \(\mathrm{KE}\) when it has slid by \(\theta\)} & \text { b. } \text{\(\frac{2 R g}{l}\left[\sin \left(\frac{l}{R}\right)+\sin \theta-\sin \left(\theta+\frac{l}{R}\right)\right]\)} \\ \hline \text { iii. } \text{The initial tangential acceleration} & \text { c. } \text{\(\frac{M R^{2} g}{l} \sin \left(\frac{l}{R}\right)\)}\\\ \hline \text { iv. } \text{The radial acceleration \(a_{r}\)} & \text { d. } \text{\(\frac{M R^{2} g}{l}\left[\sin \left(\frac{l}{R}\right)+\sin \theta-\sin \left(\theta+\frac{l}{R}\right)\right]\)} \\ \hline \end{array} $$

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