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An object moving in the \(xy\)-plane is acted on by a conservative force described by the potential-energy function \(U(x, y) = \alpha[(1/x^2) + (1/y^2)]\), where a is a positive constant. Derive an expression for the force expressed in terms of the unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\).

Short Answer

Expert verified
\( \vec{F} = \left( \frac{2\alpha}{x^3} \right) \hat{\imath} + \left( \frac{2\alpha}{y^3} \right) \hat{\jmath} \).

Step by step solution

01

Understand the Relationship between Force and Potential Energy

The force acting on an object is related to the potential energy function by the negative gradient of the potential energy. This means the force can be expressed as \( \vec{F} = -abla U(x, y) \).
02

Calculate the Partial Derivative with Respect to x

Find the partial derivative of the potential energy function \( U(x, y) = \alpha \left( \frac{1}{x^2} + \frac{1}{y^2} \right) \) with respect to \( x \). \[ \frac{\partial U}{\partial x} = \alpha \left( \frac{d}{dx} \left( \frac{1}{x^2} \right) \right) = \alpha \left( -\frac{2}{x^3} \right) = -\frac{2\alpha}{x^3}. \]
03

Calculate the Partial Derivative with Respect to y

Find the partial derivative of the potential energy function with respect to \( y \). \[ \frac{\partial U}{\partial y} = \alpha \left( \frac{d}{dy} \left( \frac{1}{y^2} \right) \right) = \alpha \left( -\frac{2}{y^3} \right) = -\frac{2\alpha}{y^3}. \]
04

Construct the Force Vector

The force vector in terms of unit vectors \( \hat{\imath} \) and \( \hat{\jmath} \) is constructed using the negative of the partial derivatives. Thus, the force \( \vec{F} \) is: \[ \vec{F} = - \left( \frac{\partial U}{\partial x} \right) \hat{\imath} - \left( \frac{\partial U}{\partial y} \right) \hat{\jmath} = \left( \frac{2\alpha}{x^3} \right) \hat{\imath} + \left( \frac{2\alpha}{y^3} \right) \hat{\jmath}. \]

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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 an essential concept in physics. It represents the stored energy of an object due to its position relative to other objects. In our example, the potential energy function is given as \(U(x, y) = \alpha \left[\frac{1}{x^2} + \frac{1}{y^2}\right]\). Here, \(\alpha\) is a constant, and the terms \(\frac{1}{x^2}\) and \(\frac{1}{y^2}\) describe how the potential energy changes with position in the \(x\)- and \(y\)-directions.
Understanding potential energy is crucial because it helps us determine the work done by or against forces. In conservative systems, like the one described in the exercise, potential energy can be converted into kinetic energy and vice versa. This interchange makes analyzing problems straightforward, without needing to account for dissipative forces like friction.
Gradient
The gradient is a vector that points in the direction of the greatest rate of increase of a function. In two dimensions, such as the \(xy\)-plane, the gradient of the potential energy \(U(x, y)\) is given by\[abla U = \left( \frac{\partial U}{\partial x} \right) \hat{\imath} + \left( \frac{\partial U}{\partial y} \right) \hat{\jmath}\]This vector provides valuable information about how the potential energy changes at each point in space.
The relationship between force and potential energy involves the negative of this gradient. For conservative forces, the force is given by \(\vec{F} = -abla U\). This means that the force points in the direction of the steepest decrease of the potential energy, effectively moving objects from high potential energy areas towards lower potential energy regions.
Partial Derivatives
Partial derivatives are a mathematical tool used to analyze functions with multiple variables. They represent the rate of change of a function concerning one variable while holding the others constant.
In this exercise, we compute the partial derivatives of the potential energy function with respect to \(x\) and \(y\):
  • For \(x\), \(\frac{\partial U}{\partial x} = -\frac{2\alpha}{x^3}\).
  • For \(y\), \(\frac{\partial U}{\partial y} = -\frac{2\alpha}{y^3}\).
These derivatives measure how the potential energy changes with small shifts in the \(x\)- or \(y\)-directions. Mastery of partial derivatives is vital because it allows us to understand how systems respond to changes in different directions, thus helping calculate forces as needed in this problem.
Unit Vectors
Unit vectors are indispensable tools for vector representation in physics. They help specify direction without regard to magnitude. In the context of the problem, the force vector is expressed using unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\).
Each unit vector has a magnitude of one and points in a specific direction:
  • \(\hat{\imath}\) represents the x-direction.
  • \(\hat{\jmath}\) represents the y-direction.
By using these vectors, we can compactly express the force \(\vec{F}\) as \[ \vec{F} = \left(\frac{2\alpha}{x^3}\right) \hat{\imath} + \left(\frac{2\alpha}{y^3}\right) \hat{\jmath}\] as derived in the solution. Using unit vectors in calculations provides a clear, concise way to convey both magnitude and direction, making complex multidimensional computations more manageable.

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Most popular questions from this chapter

The potential energy of two atoms in a diatomic molecule is approximated by \(U(r) = (a/r^{12}) - (b/r^6)\), where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) on one atom as a function of \(r\). Draw two graphs: one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r\). (b) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to \(dissociate\) it\(-\)that is, to separate the two atoms to an infinite distance apart? This is called the \(dissociation\) \(energy\) of the molecule. (d) For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is 1.13 \(\times\) 10\(^{-10}\) m and the dissociation energy is 1.54 \(\times\) 10\(^{-18}\) J per molecule. Find the values of the constants \(a\) and \(b\).

A certain spring found not to obey Hooke's law exerts a restoring force \(Fx(x) = -ax - \beta x^2\) if it is stretched or compressed, where \(\alpha\) = 60.0 N/m and \(\beta\) = 18.0 N/m2. The mass of the spring is negligible. (a) Calculate the potential-energy function U(\(x\)) for this spring. Let \(U = 0\) when \(x = 0\). (b) An object with mass 0.900 kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 m to the right (the \(+x\)-direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 m to the right of the \(x = 0\) equilibrium position?

A 0.150-kg block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is 1.20 m above the floor. The spring has force constant 1900 N/m and is initially compressed 0.045 m. The mass of the spring is negligible. The spring is released, and the block slides along the table, goes off the edge, and travels to the floor. If there is negligible friction between the block of ice and the tabletop, what is the speed of the block of ice when it reaches the floor?

A small block with mass 0.0400 kg slides in a vertical circle of radius \(R =\) 0.500 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point \(A\), the normal force exerted on the block by the track has magnitude 3.95 N. In this same revolution, when the block reaches the top of its path, point \(B\), the normal force exerted on the block has magnitude 0.680 N. How much work is done on the block by friction during the motion of the block from point \(A\) to point \(B\)?

A small block with mass 0.0500 kg slides in a vertical circle of radius \(R =\) 0.800 m on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block's path, the normal force the track exerts on the block has magnitude 3.40 N. What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

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