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The potential energy for a force field \(\vec{F}\) is given \(\%\) \(U(x, y)=\cos (x+y) .\) The force acting on a particle Il position given by coordinates \((0, \pi / 4)\) is (1) \(-\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})\) (2) \(\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})\) (3) \(\left(\frac{1}{2} \hat{i}+\frac{\sqrt{3}}{2} \hat{j}\right)\) (4) \(\left(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j}\right)\)

Short Answer

Expert verified
The correct force is option (2): \( \frac{1}{\sqrt{2}}(\hat{i} + \hat{j}) \).

Step by step solution

01

Understand the Relationship Between Force and Potential Energy

The force in a force field derived from a potential energy function \( U(x, y) \) is given by the negative gradient of the potential energy function. Mathematically, it is expressed as \( \vec{F} = -abla U(x, y) \). For potential energy \( U(x, y) = \cos(x + y) \), we will compute the gradient.
02

Compute the Gradient of the Potential Energy

The gradient \( abla U(x, y) \) is a vector composed of the partial derivatives with respect to \( x \) and \( y \). Compute these as follows:\[ \frac{\partial U}{\partial x} = -\sin(x + y) \]\[ \frac{\partial U}{\partial y} = -\sin(x + y) \]Thus, \( abla U(x, y) = (-\sin(x + y)) \hat{i} + (-\sin(x + y)) \hat{j} \).
03

Evaluate the Gradient at the Given Coordinates

Substitute \((x, y) = (0, \frac{\pi}{4}) \) into the expressions for the gradient.- \( \sin(0 + \frac{\pi}{4}) = \frac{1}{\sqrt{2}} \)Thus, the gradient at this point is:\[ abla U(0, \frac{\pi}{4}) = \left(-\frac{1}{\sqrt{2}}\right) \hat{i} + \left(-\frac{1}{\sqrt{2}}\right) \hat{j} \]
04

Determine the Force Using the Negative Gradient

The force \( \vec{F} \) is the negative of the gradient:\[ \vec{F} = -\left[ \left(-\frac{1}{\sqrt{2}}\right) \hat{i} + \left(-\frac{1}{\sqrt{2}}\right) \hat{j} \right] = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \]
05

Match the Result with Given Options

Compare the calculated force \( \vec{F} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{j}) \) with the provided choices. The correct answer is:(2) \( \frac{1}{\sqrt{2}}(\hat{i} + \hat{j}) \).

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Key Concepts

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

Gradient
The concept of a gradient is central to understanding how potential energy translates into force within a force field. Simply put, the gradient of a function gives the direction and rate of greatest increase for that function. For a potential energy function like \( U(x, y) \), the gradient is a vector field where each vector points in the direction of the steepest ascent of \( U \). This is crucial because the force acting on a particle in a force field is directly proportional to the negative of this gradient. In mathematical terms, if you have a potential energy function of \( U(x, y) = \cos(x + y) \), the force can be calculated by taking the negative gradient of \( U(x, y) \). Understanding the gradient helps in determining how the potential energy varies in space and hence how the force behaves.
Force Field
Force fields are invisible areas within which a force acts on anything that enters it. Think of them like an invisible net that affects objects moving through it. In physics, force fields like the gravitational and electromagnetic fields are types of force fields that influence objects and particles.
These fields exert force based on position within the field, which can be mapped by potential energy functions. The energy being potentially stored calculates the field's effect on a point particle within these coordinates.- A force field's force results from the negative gradient of the potential energy, which tells us that it works towards minimizing energy.- The given potential energy function \( U(x, y) = \cos(x + y) \) describes one such scenario.
This is why identifying how the potential energy varies helps in finding the field's force at specific points.
Partial Derivatives
Partial derivatives are used to find the rate at which a function changes with respect to one of its variables while keeping the other variables constant. For multivariable functions like \( U(x, y) = \cos(x + y) \), they provide information on how the function changes if only one of the variables changes.
- To calculate these for potential energy, take the derivative of the function with respect to \( x \), treating \( y \) as a constant, and vice versa. - For instance, partial derivatives of \( U(x, y) \) are: - \( \frac{\partial U}{\partial x} = -\sin(x + y) \) - \( \frac{\partial U}{\partial y} = -\sin(x + y) \)
Partial derivatives are critical in forming the components of the gradient vector. They specify how much a small change in one direction will affect the function's value, giving insight into the function's behavior in a multidimensional space.

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Most popular questions from this chapter

One end of a spring of force constant \(k_{1}\) is attached to the ceiling of an elevator. A block of mass \(1.5 \mathrm{~kg}\) is attached to the other end. Another spring of force constant \(k_{2}\) is attached to the bottom of the mass and to the floor of the elevator as shown in the figure. At equilibrium, the deformation in both the springs is equal and is \(40 \mathrm{~cm}\). If the elevator moves with constant acceleration upward, the additional deformation in both the springs is \(8 \mathrm{~cm}\). Find the elevator's acceleration \(\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)\).

A boy of mass \(m\) climbs up a conveyor belt with a constant acceleration. The speed of the belt is \(v=\sqrt{g h / 6}\) and the coefficient of friction between the boy and conveyor belt is \(\mu=\frac{5}{3 \sqrt{3}}\), The boy starts from \(A\) and moves with the maximum possible acceleration till he reaches the highest point \(B\). Work done by the boy is (1) \(\frac{5}{6} m g h\) (2) \(\frac{1}{4} m g h\) (3) \(\frac{4}{3} m g h\) (4) None of above

Mark the correct statement(s). (1) Total work done by internal forces of a system on the system is always zero. (2) Total work done by internal forces of a system on the system is sometimes zero. (3) Total work done by internal forces acting between the particles of a rigid body is always zero. (4) Total work done by internal forces acting between the particles of a rigid body is soinetimes zero.

The potential energy of a particle is determined by the expression \(U=\alpha\left(x^{2}+y^{2}\right)\), where \(\alpha\) is a positive constant. The particle begins to move from a point with coordinates \((3,3)\), only under the action of potential field force. Then its kinetic energy \(T\) at the instant when the particle is at a point with the coordinates \((1,1)\) is (1) \(8 \alpha\) (2) \(24 \alpha\) (3) \(16 \alpha\) (4) Zero

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) \(16.9 \mathrm{~J}\)

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