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Find the force corresponding to the potential energy \(U(x)=-a / x+b / x^{2}\)

Short Answer

Expert verified
The force corresponding to the potential energy \(U(x)=-\frac{a}{x} + \frac{b}{x^2}\) is \(F(x) = -\frac{a}{x^2} + 2\frac{b}{x^3}\).

Step by step solution

01

Identify the Potential Energy Function

The given potential energy function is \(U(x) = -\frac{a}{x} + \frac{b}{x^2}\). This function will be used in the following calculation.
02

Calculate the Derivative of the Potential Energy Function

The force \(F(x)\) is given by \(F(x)=-\frac{dU}{dx}\). Therefore, it requires finding the derivative of \(U(x)\). Apply the power rule of differentiation, which, for any real number n, states that the derivative of \(x^n\) is \(nx^{n-1}\). The derivative of \(U(x)\) will be \(\frac{dU}{dx}= \frac{a}{x^2}-2\frac{b}{x^3}\).
03

Calculate the Force

Now, we substitute the derivative result above into the force equation. Thus, the force \(F(x)\) on the particle at any position x will be \(F(x)=-\frac{dU}{dx}=-( \frac{a}{x^2}-2\frac{b}{x^3})= -\frac{a}{x^2} + 2\frac{b}{x^3}\).

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Key Concepts

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

Potential Energy Function
In physics, the potential energy function represents the stored energy of an object due to its position, condition, or configuration. This type of energy is associated with the forces acting upon the object and can be viewed as the capacity for doing work which is released as the object changes position. For example, gravitational potential energy increases as an object is raised higher above the ground, storing energy that will transform into kinetic energy when the object is allowed to fall.

In the given exercise, the potential energy function is expressed as a combination of terms involving the reciprocal of the position variable, x. Specifically, the function is given by:
\[\begin{equation}U(x) = -\frac{a}{x} + \frac{b}{x^2}\end{equation}\]
This equation might represent the potential energy due to different force fields or specific case scenarios in physics. The important thing to note is that the potential energy function is central to understanding how forces will act upon an object at different positions, especially when the object is subject to variable forces.
Derivative of Potential Energy
When working with potential energy in physics, it is often important to evaluate how this energy changes with respect to position. This change is precisely what the derivative of the potential energy function denotes. By finding the derivative, one can understand the rate at which potential energy is changing with position, which is directly related to the force experienced by the object.

The derivative of a potential energy function is a fundamental calculation in determining the force acting on an object in physics. According to Newton's laws, the force acting on an object is related to the gradient or slope of its potential energy curve. In mathematical terms, if you have a potential energy function, U(x), the force F(x) is the negative gradient of this function. The relationship can be expressed as: \[\begin{equation}F(x) = -\frac{dU}{dx}\end{equation}\]
In practice, calculating derivatives can vary in complexity. For our specific function, the power rule makes this process straightforward, as we'll explore in the next section.
Power Rule of Differentiation
The power rule of differentiation is an essential tool in calculus, particularly when dealing with polynomial functions. It states that for any function of the form \[\begin{equation}f(x) = x^n\end{equation}\],
where n is any real number, the derivative of that function with respect to x is \[\begin{equation}f'(x) = nx^{n-1}\end{equation}\].
In the context of our potential energy function, this rule allows us to efficienty calculate the derivative by treating each term in the function separaely. You can see this in action in the step by step solution for our exercise, where we applied the power rule to each term to find the forces correspond to the potential energy.

Applying the Power Rule

Let's look at how the power rule is applied to the potential energy function \[\begin{equation}U(x) = -\frac{a}{x} + \frac{b}{x^2}\end{equation}\].To apply the power rule, first rewrite each term:
- For \[\begin{equation}-\frac{a}{x}\end{equation}\], which is \[\begin{equation}-ax^{-1}\end{equation}\], the derivative is \[\begin{equation}-(-1)ax^{-2}\end{equation}\] or \[\begin{equation}\frac{a}{x^2}\end{equation}\].
- For \[\begin{equation}\frac{b}{x^2}\end{equation}\], which is \[\begin{equation}bx^{-2}\end{equation}\], the derivative is \[\begin{equation}-2bx^{-3}\end{equation}\].
Combining these, we get the derivative of the potential energy function, crucial for finding the force as outlined in step 2 of the solution.
Force in Physics
Force is a fundamental concept in physics, representing the influence that changes the motion of an object. According to Newton's second law, it is also defined as the mass of an object multiplied by its acceleration, and expressed in the metric system in newtons (N). Forces can be caused by phenomena such as gravity, magnetism, or any interaction that can change an object's speed or direction of movement.

In the context of our exercise, the force determined from the potential energy function relates to conservative forces, which are path-independent and solely a function of position. The calculation includes taking the negative derivative of the potential energy function because forces in conservative fields are always directed towards lowering the potential energy of the system.

The negative sign ensures that the force vector points in the direction of the greatest decrease in potential energy. In simpler terms, a system will naturally move in the direction where it can lower its potential energy, which underlies the entire concept of stability in physical systems.

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Most popular questions from this chapter

A girl on a skateboard (total mass of 40 kg) is moving at a speed of \(10 \mathrm{m} / \mathrm{s}\) at the bottom of a long ramp. The ramp is inclined at \(20^{\circ}\) with respect to the horizontal. If she uavels \(14.2 \mathrm{m}\) upward along the ramp before stopping. what is the net frictional force on her?

A block of mass 300 g is attached to a spring of spring constant \(100 \mathrm{N} / \mathrm{m}\). The other end of the spring is attached to a support while the block rests on a smooth horizontal table and can slide freely without any friction. The block is pushed horizontally till the spring compresses by \(12 \mathrm{cm}\) and then the block is released from rest. (a) How much potential energy was stored in the block-spring support system when the block was just released? (b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched. (c) Determine the speed of the block when it has traveled a distance of 20 \(\mathrm{cm}\) from where it was released.

A block of mass 200 g is attached at the end of a massless spring of spring constant \(50 \mathrm{N} / \mathrm{m}\). The other end of the spring is attached to the ceiling and the mass is released at a height considered to be where the gravitational potential energy is zero. (a) What is the net potential energy of the block at the instant the block is at the lowest point? (b) What is the net potential energy of the block at the midpoint of its descent? (c) What is the speed of the block at the midpoint of its descent?

What is the physical meaning of a non-conservative force?

A child (32 kg) jumps up and down on a trampoline. The trampoline exerts a spring restoring force on the child with a constant of \(5000 \mathrm{N} / \mathrm{m}\). At the highest point of the bounce, the child is \(1.0 \mathrm{m}\) above the level surface of the trampoline. What is the compression distance of the trampoline? Neglect the bending of the legs or any transfer of energy of the child into the trampoline while jumping.

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