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CALC In an experiment, one of the forces exerted on a proton is \(\overrightarrow{\boldsymbol{F}}=-\alpha x^{2} \hat{\imath},\) where \(\alpha=12 \mathrm{~N} / \mathrm{m}^{2}\). (a) How much work does \(\boldsymbol{F}\) do when the proton moves along the straight-line path from the point \((0.10 \mathrm{~m}, 0)\) to the point \((0.10 \mathrm{~m}, 0.40 \mathrm{~m}) ?\) (b) Along the straight-line path from the point \((0.10 \mathrm{~m}, 0)\) to the point \((0.30 \mathrm{~m}, 0) ?\) (c) Along the straight-line path from the point \((0.30 \mathrm{~m}, 0)\) to the point \((0.10 \mathrm{~m}, 0) ?\) (d) Is the force \(\vec{F}\) conservative? Explain. If \(\boldsymbol{F}\) is conservative, what is the potential-energy function for it? Let \(U=0\) when \(x=0\).

Short Answer

Expert verified
The work done in part (a) is 0 because the force is perpendicular to the displacement. The work done for part (b) and part (c) can be calculated by integrating the force along the path of displacement. Whether the force is conservative can be determined comparing the work done in parts (b) and (c). If force is conservative, the potential energy function can be obtained by evaluating the integral of force with respect to displacement.

Step by step solution

01

Understanding The Force And Movement

We are given that a force is acting on the proton given by \(\overrightarrow{F} = -\alpha x^2 \hat{i}\), where \(\alpha = 12 N/m^2\). The force is only acting along the x direction since it is in terms of i-hat. Moreover, the proton is moving along the x and y directions in various parts of the problem.
02

Calculate Work Done For Part (a)

In part (a), the proton moves from the point (0.10 m, 0) to the point (0.10 m, 0.40 m). Since the force is only in the x direction and the motion of the proton is only in the y direction in this part, they are perpendicular to each other. The work done by a force two vectors which are perpendicular to each other is zero. Therefore, the work done in part (a) is zero.
03

Calculate Work Done For Part (b)

In part (b), the proton moves from the point (0.10 m, 0) to the point (0.30 m, 0). Now we are given a displacement along the x-direction. The work done by the force during this movement will be the integral of force \(\overrightarrow{F}\) along the path of the proton. Since \(\overrightarrow{F} = -\alpha x^2 \hat{i}\), work done will be \(\int_{0.10}^{0.30} (-\alpha x^2) dx\). Simplifying this integral, we get the work done by the force in part (b).
04

Calculate Work Done For Part (c)

In part (c), the proton moves back from the point (0.30 m, 0) to the point (0.10 m, 0). The work done by the force during this movement will be the integral of force \(\overrightarrow{F}\) along the path of the proton. We will integrate from 0.30 to 0.10, \(\int_{0.30}^{0.10} (-\alpha x^2) dx\). Simplifying this integral, we get the work done by the force in part (c).
05

Checking If The Force \(\overrightarrow{F}\) Is Conservative

A force is conservative if the work done by it is independent of the path taken. It only depends on the initial and final states. By comparing the work done in part (b) and part (c), we can say whether the force is conservative or not. If the absolute work done is the same for both parts, the force is conservative, otherwise it's not.
06

Calculate Potential Energy Function

If the force is conservative, the potential energy function \(U\) associated with the force can be calculated using the equation \(U = - \int \overrightarrow{F} \cdot dr\), where \(dr\) is the displacement. We assume 'U = 0' at 'x = 0' in this case. As \(dr = dx \hat{i}\) and \(\overrightarrow{F} = -\alpha x^2 \hat{i}\), the potential energy U(x) can be found by evaluating \(U(x) = - \int (-\alpha x^2) dx\). This will give the potential energy function for the force \(F\).

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Key Concepts

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

Conservative Force
When we talk about a conservative force, we refer to a type of force where the work done in moving an object between two points is solely dependent on the initial and final positions of the object, not on the path taken. This characteristic is essential for a variety of physical phenomena and theoretical calculations, as it simplifies many complex scenarios.

A classic example of a conservative force is gravity; whether you drop an object straight down or slide it down a ramp, the work done by gravity only depends on the height difference. Similarly, in the textbook problem provided, we're tasked with determining whether the force exerted on a proton, indicated by \( \-\alpha x^{2} \hat{\imath} \), is a conservative force. This is done by examining the work done by the force over different paths and checking for consistency.

If after various calculations, the work done is found to be path-independent, we can declare the force to be conservative. Then, a significant implication arises: we can associate a potential energy function with this force, fundamental for understanding energy conservation and storage within the system.
Potential Energy Function
The potential energy function is a powerful concept used to analyze the energy stored in a system due to the position of an object within a force field. For conservative forces, this function is instrumental in computing the work done without performing complex integrals every time an object moves. In simpler terms, potential energy represents the latent capability to do work based on an object's position or configuration.

Returning to our example, if the force \( \vec{F} \) is conservative, the associated potential energy function, denoted as \( U(x) \), can be derived mathematically by calculating the negative integral of the force over distance. This function is a measure of the energy that would be released (or required) as an object moves from one point to another within the force field, and it's essential for understanding the total energy dynamics of the system.
Force Exerted on a Proton
In our context, a force exerted on a proton is described mathematically by the expression \( \-\alpha x^{2} \hat{\imath} \). This indicates that the force has a magnitude that varies with the square of the proton's position along the x-axis, and it points in the negative x-direction. This specific form suggests that as the proton moves further from the origin, the force exerted on it grows stronger.

Understanding the nature of this force is crucial for solving the textbook problem, where we're asked to calculate the work done as the proton moves along different paths. Notice that the force depends on the displacement in the x-axis, making it vital to consider the direction of motion when calculating work. For instance, when the proton moves solely in the y-axis, no work is done by this force, because work is only done by the component of force in the direction of displacement.
Work-Energy Principle
The work-energy principle is foundational in physics and asserts that the work done by all the forces acting on an object results in a change in the object’s kinetic energy. Applied to our scenario, analyzing how the force \( \vec{F} \) affects the proton's movement requires utilizing this principle.

The principle can be straightforwardly stated as \( W = \Delta K \), where \( W \) represents the net work done on the object, and \( \Delta K \) is the change in its kinetic energy. When the net work done is zero, as in the case when the proton moves perpendicular to the force, there is no change in kinetic energy. On the other hand, when work is done, we expect to see a corresponding change in the proton's kinetic energy. Through this conceptual lens, we can precisely analyze the relationship between force, motion, and energy — a fundamental trio in dynamics.

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Most popular questions from this chapter

The Great Sandini is a \(60 \mathrm{~kg}\) circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of \(1100 \mathrm{~N} / \mathrm{m}\) that he will compress with a force of \(4400 \mathrm{~N}\). The inside of the gun barrel is coated with Teflon, so the average friction force will be only \(40 \mathrm{~N}\) during the \(4.0 \mathrm{~m}\) he moves in the barrel. At what speed will he emerge from the end of the barrel, \(2.5 \mathrm{~m}\) above his initial rest position?

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BIO Human Energy vs. Insect Energy. For its size, the common flea is one of the most accomplished jumpers in the animal world. A 2.0 -mm-long, \(0.50 \mathrm{mg}\) flea can reach a height of \(20 \mathrm{~cm}\) in a single leap. (a) Ignoring air drag, what is the takeoff speed of such a flea? (b) Calculate the kinetic energy of this flea at takeoff and its kinetic energy per kilogram of mass. (c) If a \(65 \mathrm{~kg}, 2.0-\mathrm{m}\) -tall human could jump to the same height compared with his length as the flea jumps compared with its length, how high could the human jump, and what takeoff speed would the man need? (d) Most humans can jump no more than \(60 \mathrm{~cm}\) from a crouched start. What is the kinetic energy per kilogram of mass at takeoff for such a \(65 \mathrm{~kg}\) person? (e) Where does the flea store the energy that allows it to make sudden leaps?

CALC A small object with mass \(m=0.0900 \mathrm{~kg}\) moves along the \(+x\) -axis. The only force on the object is a conservative force that has the potential-energy function \(U(x)=-\alpha x^{2}+\beta x^{3},\) where \(\alpha=2.00 \mathrm{~J} / \mathrm{m}^{2}\) and \(\beta=0.300 \mathrm{~J} / \mathrm{m}^{3} .\) The object is released from rest at small \(x .\) When the object is at \(x=4.00 \mathrm{~m},\) what are its (a) speed and (b) acceleration (magnitude and direction)? (c) What is the maximum value of \(x\) reached by the object during its motion?

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