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Chapter 7: Problem 35

EALE A force parallel to the \(x\) -axis acts on a particle moving along the \(x\) -axis. This force produces potential energy \(U(x)\) given by \(U(x)=\alpha x^{4},\) where \(\alpha=1.20 \mathrm{J} / \mathrm{m}^{4} .\) What is the force (magnitude and direction) when the particle is at \(x=-0.800 \mathrm{m}\) ?

Short Answer

Expert verified
The force magnitude is 2.46 N, towards the positive x-direction.

Step by step solution

01

Understand the relationship between force and potential energy

The force acting on a particle is related to the potential energy by the negative gradient of the potential energy function. In one dimension, this is expressed as: \( F(x) = -\frac{dU}{dx} \).
02

Differentiate the potential energy function

Given the potential energy \( U(x) = \alpha x^4 \), we differentiate \( U(x) \) with respect to \( x \) to find \( \frac{dU}{dx} \). Thus, \( \frac{dU}{dx} = 4\alpha x^3 \).
03

Substitute the values to find the force

Substitute \( \alpha = 1.20 \mathrm{J/m^4} \) and \( x = -0.800 \mathrm{m} \) into the expression for force. This gives: \[ F(x) = -4 \times 1.20 \times (-0.800)^3 \].
04

Calculate the force

Compute \((-0.800)^3\) which is \(-0.512\). Multiply by \(4 \times 1.20\) to find: \[ F(x) = -4 \times 1.20 \times -0.512 = 2.4576 \text{ N} \approx 2.46 \text{ N}\].
05

Determine the direction of the force

Since the force is calculated to be positive, it indicates the direction of the force is in the positive \( x \)-axis direction.

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Key Concepts

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

Force
In physics, force is a push or pull on an object that causes it to change its velocity or direction. When a force is applied along the \(x\)-axis, it essentially acts in the horizontal direction. For this context, we focus on how force relates to potential energy.

Potential energy refers to the stored energy in a system due to its position or configuration. In this problem, the force is related to the potential energy by the formula \( F(x) = -\frac{dU}{dx} \). This equation indicates that the force is the negative derivative of the potential energy function \(U(x)\).

Key points to remember:
  • Force affects the motion and position of a particle.
  • Force and potential energy are inherently related through differentiation.
  • In this problem, a positive force means it's directed along the positive \(x\)-axis.
Differentiation
Differentiation is a fundamental concept in calculus that helps us understand how a function changes at a particular point. It is the process of finding the derivative of a function.

In this exercise, differentiation enables us to find out how the potential energy changes as the position \(x\) changes. To differentiate \( U(x) = \alpha x^4 \), we use the power rule, which states that if \( f(x) = x^n \), then \( \frac{df}{dx} = nx^{n-1} \). Applying this rule, the derivative of the potential energy function is \( \frac{dU}{dx} = 4\alpha x^3 \).

Points to note:
  • Differentiation gives us the slope or rate of change of a function.
  • The derivative is crucial in assessing how potential energy transforms into kinetic energy through force application.
  • Understanding and applying derivative rules facilitates computations in physics problems.
Gradient
In this context, the gradient refers to the derivative of the potential energy function with respect to position. It is commonly called the slope and indicates how the function changes over space.

The gradient is essential because it tells us the direction and rate at which the potential energy changes, which directly influences force. When we say that force is the "negative gradient" of potential energy, it means the force points in the direction where the potential energy decreases most rapidly.

Important considerations:
  • The gradient is the measure of change in the potential energy with respect to position.
  • A negative gradient signifies that the force and the direction of energy increment are opposite.
  • Analyzing the gradient helps in understanding the stability and movement of particles in a field.

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Most popular questions from this chapter

CALC An object has several forces acting on it. One force is \(\vec{\boldsymbol{F}}=\alpha x y \hat{\boldsymbol{r}},\) a force in the \(x\) -direction whose magnitude depends on the position of the object. (Sce Problem \(6.98 . )\) The constant is \(\alpha=2.00 \mathrm{N} / \mathrm{m}^{2}\) . The object moves along the following path: \((1)\) It starts at the origin and moves along the \(y\) -axis to the point \(x=0\) , \(y=1.50 \mathrm{m} ;(2)\) it moves parallel to the \(x\) -axis to the point \(x=1.50 \mathrm{m}, y=1.50 \mathrm{m} ;(3)\) it moves parallel to the \(y\) -axis to the point \(x=1.50 \mathrm{m}, y=0 ;(4)\) it moves parallel to the \(x\) -axis back to the origin. (a) Sketch this path in the \(x y\) -plane. (b) Calculate the work done on the object by \(\vec{F}\) for each leg of the path and for the complete round trip. (c) Is \(\vec{F}\) conservative or nonconservative? Explain.

CALC The potential energy of a pair of hydrogen atoms separated by a large distance \(x\) is given by \(U(x)=-C_{6} / x^{6},\) where \(C_{6}\) is a positive constant. What is the force that one atom exerts on the other? Is this force attractive or repulsive?

EP Pendulum. A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 \(\mathrm{m}\) to form a pendulum. The pendulum is swinging so as to make a maximum angle of \(45^{\circ}\) with the vertical. Air resistance is negligible. (a) What is the speed of the rock when the string passes through the vertical position? (b) What is the tension in the string when it makes an angle of \(45^{\circ}\) with the vertical? (c) What is the tension in the string as it passes through the vertical?

A basket of negligible weight hangs from a vertical spring scale of force constant 1500 \(\mathrm{N} / \mathrm{m}\) . (a) If you suddenly put a \(3.0-\mathrm{kg}\) adobe brick in the basket, find the maximum distance that the spring will stretch. (b) If, instead, you release the brick from 1.0 \(\mathrm{m}\) above the basket, by how much will the spring stretch at its maximum elongation?

BIO How High Can We Jump? The maximum height a typical human can jump from a crouched start is about 60 \(\mathrm{cm}\) . By how much does the gravitational potential energy increase for a 72 -kg person in such a jump? Where does this energy come from?
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