Potential Energy vs. Position Graphs

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Homework Help Overview

The discussion revolves around a problem involving potential energy and its relationship to force, specifically focusing on calculating the x-component of the force on a particle at various positions based on a given potential energy graph.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the slope of the potential energy graph and the force, questioning the implications of a constant slope in a specific range. There are attempts to clarify the correct application of the formula for force derived from potential energy.

Discussion Status

Some participants have provided guidance on the interpretation of the slope and its implications for force, suggesting that the original poster may have misapplied the distance in their calculations. The discussion reflects a collaborative effort to clarify misunderstandings without reaching a definitive conclusion.

Contextual Notes

There is an emphasis on checking units and understanding the relationship between average and instantaneous rates of change in the context of the problem. The original poster expresses confusion about their calculations, indicating a need for further exploration of the concepts involved.

sunnnystrong
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Homework Statement


A particle has the potential energy shown in the figure. (Figure 1) What is the
x-component of the force on the particle at x =5, 15, 25, and 35 cm?

I'm stuck on finding the Fx at 25 & 35 cm.

Homework Equations



F = -du/dx
100 N = 1J/cm

The Attempt at a Solution



Slope throughout 20-40 cm is a constant positive .5.

So at 25 cm it should be :
F = -.5/25cm * 100 = -2 N

This is incorrect. I'm really confused?
 

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Hey! You're spot on that the slope is constant throughout the 20cm-40cm range which means the average rate of change in that region is equal to the instantaneous rate of change at all points in the region. You also correctly pointed out that the expression for force is the negative or the rate of change of energy. So if the rate of change (slope) is constant in that region, what does that imply about the force at all points in the region? If you think about that question it should reveal that your problem lies in dividing by 25 cm. It's also a good idea to check your units if you run into these problems. Dividing the slope by a distance unit doesn't give you a Newton does it?
 
Supercool_ed said:
Hey! You're spot on that the slope is constant throughout the 20cm-40cm range which means the average rate of change in that region is equal to the instantaneous rate of change at all points in the region. You also correctly pointed out that the expression for force is the negative or the rate of change of energy. So if the rate of change (slope) is constant in that region, what does that imply about the force at all points in the region? If you think about that question it should reveal that your problem lies in dividing by 25 cm. It's also a good idea to check your units if you run into these problems. Dividing the slope by a distance unit doesn't give you a Newton does it?
Hello,
so i calculated the force applied at 25cm to be -2 N which is incorrect :( I'm just really confused as to what is wrong as i feel like i derived it correctly
 
sunnnystrong said:
Hello,
so i calculated the force applied at 25cm to be -2 N which is incorrect :( I'm just really confused as to what is wrong as i feel like i derived it correctly
Let's go back to the basic equation. So, F=-dU/dx. If the slope is constant then this can be written as F=-[U(40)-U(20)]/[40-20] (then multiply by 100 to convert from cm to m) You could use any points in the interval, but will get the same value because, as you pointed out, the slope is constant (the same over whole interval)
 
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Supercool_ed said:
Let's go back to the basic equation. So, F=-dU/dx. If the slope is constant then this can be written as F=-[U(40)-U(20)]/[40-20] (then multiply by 100 to convert from cm to m) You could use any points in the interval, but will get the same value because, as you pointed out, the slope is constant (the same over whole interval)

Ohh okay thank you hahah i get it now. The force will be constant everywhere that makes more sense now
 

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