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The potential energy of a system of two particles separated by a distance \(r\) is given by \(U(r)=A / r,\) where \(A\) is a constant. Find the radial force \(\mathbf{F}_{r}\) that each particle exerts on the other.

Short Answer

Expert verified
The radial force that each particle exerts on the other is \(F = \frac{A}{r^{2}}\).

Step by step solution

01

Identify Given Variables

The potential energy \(U(r)\) of a system of two particles separated by a distance \(r\) is given by the equation \(U(r)=\frac{A}{r},\) where \(A\) is a constant.
02

Remember the relationship between force and potential energy

The force exerted by each particle on the other can be described by the relationship \(F = -\frac{dU}{dr}\). This law states that the force is equal to the negative derivative of the potential energy.
03

Apply the relationship to the given equation

Using the equation \(F = -\frac{dU}{dr}\), substitute \(U(r)\) and differentiate with respect to \(r\). Hence, \(F = -\frac{d}{dr}\left(\frac{A}{r}\right)\).
04

Solve the derivative

Upon differentiating \(\frac{A}{r}\) with respect to \(r\), we get \(F = -A\left(-\frac{1}{r^{2}}\right)\). The derivative of \(\frac{A}{r}\) with respect to \(r\) is \(-\frac{A}{r^{2}}\). Remember to keep the negative sign from the original equation.
05

Simplify the equation

After simplification we get \(F = \frac{A}{r^{2}}\). This is now simplified as the radial force that each particle exerts on the other.

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Key Concepts

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

Radial Force
When two particles are interacting through a force that depends only on their separation distance, like the potential energy provided in the exercise, we describe the force as radial. Radial force means that the force acts along the line joining the centers of the particles. It's important because it influences how particles move and interact.
In our context, the given potential energy function is based on the distance between the particles. To determine the radial force exerted by each particle on the other, we recognize that the force is not acting in a random direction but specifically along the radius vector connecting them.
  • The magnitude of this force is connected directly to their distance, decreasing rapidly as they move further apart due to the inverse-square relationship after differentiation.
  • Since we're dealing with idealized particles in the exercise, the radial force simplifies calculations and helps model how real forces, like gravitational or electrostatic interactions, might work over small scales.
Understanding radial forces helps explain anything from the orbits of planets to the behavior of electrons around a nucleus.
Derivative of Potential Energy
Taking the derivative of potential energy with respect to distance is key in determining the force between two particles. When potential energy depends on distance, like in this exercise, the force can be found by taking the derivative of the potential energy function.
The potential energy provided is a function of distance, expressed as \(U(r) = \frac{A}{r}\). To find the force, we differentiate this function with respect to \(r\). Here's how:
  • The derivative of \(\frac{A}{r}\) with respect to \(r\) is \(-\frac{A}{r^2}\). It indicates how the potential energy changes as the two particles move closer or further apart.
  • The negative sign indicates direction: the force tends to move in the direction that reduces potential energy, basically trying to "flatten" the energy difference.
In essence, finding this derivative allows us to transform information about potential energy into tangible force values, enabling predictions about motion and interactions.
Force and Potential Energy Relationship
The relationship between force and potential energy is central to understanding dynamics in physics. At its core, this relationship tells us how the conservative forces acting between particles influence their movement.
Force is the derivative of potential energy with an added negative sign, expressed as \(F = -\frac{dU}{dr}\). This equation highlights several elements:
  • The potential energy function gives a landscape of energy levels between particle positions.
  • Differentiating this function provides information about how steep or flat this energy landscape is—a steeper slope means a stronger force.
  • The negative sign indicates that forces act in a way to reduce potential energy, effectively pulling objects towards lower energy states.
By leveraging this relationship, we gain insight into how entities interact under the influence of forces such as gravity and electromagnetism. It underpins not only simple systems like the one in the exercise but also complex systems across physics.

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Most popular questions from this chapter

At time \(t_{i},\) the kinetic energy of a particle is \(30.0 \mathrm{J}\) and the potential energy of the system to which it belongs is \(10.0 \mathrm{J}\) At some later time \(t\), the kinetic energy of the particle is 18.0 J. (a) If only conservative forces act on the particle, what are the potential energy and the total energy at time \(t_{f} ?\) (b) If the potential energy of the system at time \(t_{f}\) is \(5.00 \mathrm{J},\) are there any nonconservative forces acting on the particle? Explain.

A potential-energy function for a two-dimensional force is of the form \(U=3 x^{3} y-7 x\). Find the force that acts at the point \((x, y)\).

A single conservative force acting on a particle varies as \(\mathbf{F}=\left(-A x+B x^{2}\right) \hat{\mathbf{i}} N,\) where \(A\) and \(B\) are constants and \(x\) is in meters. (a) Calculate the potential-energy function \(U(x)\) associated with this force, taking \(U=0\) at \(x=0 .\) (b) Find the change in potential energy and the change in kinetic energy as the particle moves from \(x=2.00 \mathrm{m}\) to \(x=3.00 \mathrm{m}\).

A simple pendulum, which we will consider in detail in Chapter \(15,\) consists of an object suspended by a string. The object is assumed to be a particle. The string, with its top end fixed, has negligible mass and does not stretch. In the absence of air friction, the system oscillates by swinging back and forth in a vertical plane. If the string is \(2.00 \mathrm{m}\) long and makes an initial angle of \(30.0^{\circ}\) with the vertical, calculate the speed of the particle (a) at the lowest point in its trajectory and (b) when the angle is \(15.0^{\circ}\).

A \(70.0-\mathrm{kg}\) diver steps off a \(10.0-\mathrm{m}\) tower and drops straight down into the water. If he comes to rest \(5.00 \mathrm{m}\) beneath the surface of the water, determine the average resistance force exerted by the water on the diver.

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