/*! 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 4 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 \(20 \mathrm{~J}\). Determine the equation of \(F(x)\) as a function of \(x\). (1) \(F=2+x\) (2) \(F=2+3 x\) (3) \(F=2(2-x)\) (4) \(F=3+2 x\)

Short Answer

Expert verified
The correct equation for the force is (3) \(F=2(2-x)\).

Step by step solution

01

Understanding the Potential Energy Relation

The potential energy as a function of position is given by \(U(x) = 20 + (x - 2)^2\). The conservative force \(F(x)\) can be derived from the potential energy using the relation \(F(x) = -\frac{dU}{dx}\).
02

Calculate the Derivative of Potential Energy

We need to find \(-\frac{d}{dx}[20 + (x - 2)^2]\). First, differentiate the term \((x - 2)^2\) with respect to \(x\), which gives \(\frac{d}{dx}(x - 2)^2 = 2(x - 2)\). The constant \(20\) has a derivative of zero.
03

Determine the Force Function

Substituting the derivative back, we have \(F(x) = -\frac{dU}{dx} = -2(x - 2)\). Simplify this to get \(F(x) = -2x + 4\), which matches option (3) \(F = 2(2-x)\).
04

Confirmation with Given Options

Among the given options, the expression \(2(2-x)\) simplifies to \(4-2x\), which is indeed the same as \(-2x+4\), confirming that option (3) is correct.

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Key Concepts

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

Potential Energy
In physics, potential energy is the energy stored within a system due to the position or arrangement of its components. Here, it is expressed as a function of position along the x-axis:
  • \(U(x) = 20 + (x-2)^2\)
This equation tells us how the potential energy varies with the position, \(x\).
This specific form shows that as you move away from \(x = 2\), the potential energy increases due to the squared term \((x-2)^2\), which describes a parabolic shape, centered at \(x=2\) where potential energy is minimized.
Conservative forces, such as the one we're dealing with here, are forces where the work done only depends on the initial and final positions, not the path taken. The potential energy can help us better understand how this force might act on the particle.
Kinetic Energy
Kinetic energy is the energy of motion. The formula for kinetic energy in one dimension is:
  • \(KE = \frac{1}{2}mv^2\)
where \(m\) is mass and \(v\) is velocity.
In our problem, a 1.0-kg particle has a kinetic energy of 20 Joules at \(x = 5.0\, m\).
This gives insight into the particle's speed at this point: as kinetic energy and potential energy convert into one another in conservative systems, the particle moves faster or slower depending on its energy configuration.
Derivative of Potential Energy
The derivative of potential energy with respect to position helps us find the force acting on the particle. This is derived from the principle that the force exerted by a conservative field is the negative gradient (or derivative) of the potential energy.
We calculate it as follows:
  • Find the derivative of \(U(x) = 20 + (x - 2)^2\).
  • The derivative of the constant \(20\) is \(0\).
  • The derivative of \((x - 2)^2\) is \(2(x - 2)\).
As per the relation, the force is:
  • \(F(x) = -\left(2(x - 2)\right)\)
This negative sign is crucial as it indicates that the force opposes the change in potential energy as expected in conservative forces.
Force Function
The force function describes how force varies with position, derived from the potential energy function. Using the derivative we discussed:
  • \(F(x) = -2(x - 2)\)
This simplifies to:
  • \(F(x) = -2x + 4\)
This force function shows that the force depends linearly on the position \(x\), with a negative slope.
It tells us that the force is strongest when \(x = 0\) and decreases towards \(x = 2\). Beyond this point, force acts in the opposite direction pulling the particle back towards \(x = 2\).
Among the given options, we verified that option (3) \(F = 2(2-x)\) simplifies to the same expression \(-2x + 4\), confirming it as the correct function.
This aligns with conservative forces where potential energy changes are directly linked to the forces countering or aiding motion.

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Most popular questions from this chapter

A small block of mass \(2 \mathrm{~kg}\) is kept on a rough inclined surface of inclination \(\theta=30^{\circ}\) fixed in a lift. The lift goes up with a uniform speed of \(1 \mathrm{~ms}^{-1}\) and the block does not slide relative to the inclined surface. The work done by the force of friction on the block in a time interval of \(2 \mathrm{~s}\) is (1) Zero (2) \(9.8 \mathrm{~J}\) (3) 294 I (4) \(16.9 \mathrm{~J}\)

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 \(20 \mathrm{~J}\). The maximum and minimum values of \(x\), respectively, are (1) \(7.38 \mathrm{~m},-3.38 \mathrm{~m}\) (2) \(6.38 \mathrm{~m},-4.38 \mathrm{~m}\) (3) \(7.38 \mathrm{~m},-2.38 \mathrm{~m}\) (4) \(6.38 \mathrm{~m},-2.38 \mathrm{~m}\)

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 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 block of \(4 \mathrm{~kg}\) mass starts at rest and slides a distance \(d\) down a friction less incline (angle \(30^{\circ}\) ) where it runs into a spring of negligible mass. The block slides an additional \(25 \mathrm{~cm}\) before it is brought to rest momentarily by compressing the spring. The force constant of the spring is \(400 \mathrm{~N} \mathrm{~m}^{-1}\), The value of \(d\) is (take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\) ) (1) \(25 \mathrm{~cm}\) (2) \(37.5 \mathrm{~cm}\) (3) \(62.5 \mathrm{~cm}\) (4) None of the above

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