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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 \(\alpha =\) 0.630 J/m\(^4\). What is the force (magnitude and direction) when the particle is at \(x = -0.800\) m?

Short Answer

Expert verified
The force is 1.293 N in the positive x-direction.

Step by step solution

01

Understand the relationship between force and potential energy

The force exerted by a potential energy field is related to the potential energy by the negative gradient. For a one-dimensional case like this, the force \( F(x) \) can be found using the formula \( F(x) = -\frac{dU}{dx} \). This means that the force is the negative derivative of the potential energy with respect to \( x \).
02

Calculate the derivative of the potential energy function

First, we need to find the derivative of the potential energy function \( U(x) = \alpha x^4 \) with respect to \( x \). Applying the power rule of differentiation, \( \frac{d}{dx}[x^n] = nx^{n-1} \), we get:\[ \frac{dU}{dx} = 4\alpha x^3 \]
03

Substitute the values into the derivative

Substitute \( \alpha = 0.630 \) J/m\(^4\) and \( x = -0.800 \) m into the derivative formula:\[ \frac{dU}{dx} = 4 \times 0.630 \times (-0.800)^3 \]
04

Calculate the value of the derivative at \( x = -0.800 \) m

Calculate the expression from the previous step:\((-0.800)^3 = -0.512\)\(4 \times 0.630 \times (-0.512) = -1.293 \) J/m
05

Determine the force and its direction

The force \( F(x) \) is the negative of the derivative of the potential energy:\[ F(x) = -\frac{dU}{dx} = -(-1.293) = 1.293 \text{ N} \]Since the result is positive, the force is in the positive \( x \)-direction.

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Key Concepts

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

Force Calculation
To solve for the force acting on a particle along the x-axis due to potential energy, we utilize the relationship between potential energy and force. The force can be derived from the potential energy function using its negative gradient. In simple terms, this involves finding the derivative of the potential energy with respect to position, and then taking the negative of that derivative. This resulting force has both magnitude and direction, giving you complete information about how the particle is influenced by the potential energy field.

Here's a quick step-by-step overview:
  • Identify the potential energy function, such as \( U(x) = \alpha x^4 \).
  • Compute the derivative \( \frac{dU}{dx} \) to determine how potential energy changes with position.
  • The force is then \( F(x) = -\frac{dU}{dx} \).
  • Evaluate this expression at the given position to find the force's magnitude and direction.
Understanding this relationship is crucial because it allows you to predict how the potential energy influences the motion of particles in various fields, both in physics and engineering applications.
Gradient and Derivative
The concepts of gradient and derivative are foundational in calculus, especially when exploring physical phenomena involving force and motion. The derivative measures how a function changes as its input changes. For a one-dimensional function such as potential energy, the derivative represents the rate at which the potential energy changes with respect to position.

In the case of potential energy \( U(x) = \alpha x^4 \), the derivative \( \frac{dU}{dx} \) gives us a rate of change of potential energy at any position \( x \). To fully understand the impact of these changes, consider:
  • The gradient in this context translates to the force's effect on the particle.
  • A positive derivative means increasing potential energy, while a negative derivative means decreasing energy.
  • The direction of the force is determined by the sign of the derivative; negative gradients indicate forces in the opposite direction.

    Calculating the derivative involves applying the power rule, where for a function \( x^n \), the derivative is \( nx^{n-1} \). Understanding how these changes apply in one dimension helps predict both the motion and behavior of particles affected by various forces.
Power Rule of Differentiation
The power rule is a basic but powerful tool in calculus, essential for finding derivatives of polynomial functions, which often represent physical quantities like potential energy. According to this rule, if you have a function \( f(x) = x^n \), its derivative is \( f'(x) = nx^{n-1} \). Simply multiply the power by the coefficient of \( x \) and then decrease the power by one.

Applying this rule to the potential energy function \( U(x) = \alpha x^4 \):
  • Differentiate using the power rule: \( \frac{d}{dx}[\alpha x^4] = 4\alpha x^3 \).
  • Substitute the constant \( \alpha \) after differentiation if necessary.
  • This derivative gives the rate of change of potential energy, crucial for calculating force.

    Using the power rule not only simplifies calculations but also proves to be an efficient method for understanding complex physical relationships, making it invaluable in fields such as physics and engineering.
Mastering this rule enables solving a variety of real-world problems easily, enhancing both your mathematical and analytical skills.

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Most popular questions from this chapter

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 spring of negligible mass has force constant \(k =\) 800 N/m. (a) How far must the spring be compressed for 1.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then lay a 1.60-kg book on top of the spring and release the book from rest. Find the maximum distance the spring will be compressed.

The \(food \, calorie\), equal to 4186 J, is a measure of how much energy is released when the body metabolizes food. A certain fruitandcereal bar contains 140 food calories. (a) If a 65kg hiker eats one bar, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes into increasing gravitational potential energy? (b) If, as is typical, only 20% of the food calories go into mechanical energy, what would be the answer to part (a)? (\(Note\): In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the body eliminates the rest. Metabolic efficiency varies considerably from person to person.)

Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 m that makes an angle of 45\(^\circ\) with the vertical, steps off his tree limb, and swings down and then up to Jane's open arms. When he arrives, his vine makes an angle of 30\(^\circ\) with the vertical. Determine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan's speed just before he reaches Jane. Ignore air resistance and the mass of the vine.

You are asked to design a spring that will give a 1160-kg satellite a speed of 2.50 m/s relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00g. The spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will all be negligible. (a) What must the force constant of the spring be? (b) What distance must the spring be compressed?

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