Solving Energy of a System - Plotting Graphs & Calculating Cosθ

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Homework Help Overview

The discussion revolves around understanding the energy of a system, specifically through the context of plotting graphs related to force and calculating the cosine of the angle between force and displacement vectors. Participants are exploring concepts related to work done, potential energy, and the implications of force being a function of position.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss methods for calculating work done using graphical representation and integration. Questions arise regarding the interpretation of potential energy changes and the significance of the angle between force and displacement vectors.

Discussion Status

The discussion is active with participants questioning assumptions about potential energy and the relationship between force and displacement. Some guidance has been offered regarding the calculation of work and the definition of cosθ, but multiple interpretations of the problem are being explored.

Contextual Notes

There are indications of confusion regarding the definitions of potential energy and the specific forces involved in the problem. Participants are also addressing the implications of one-dimensional motion and the role of gravity in their analyses.

haha1234
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I have tried to solve this two question.
For question 13, I've plotted a graph.Is it correct?
But I can't find the answer by calculating the area under the line.

For question 11,What is the cosθ:cry:
 

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You are plotting F(x) = 2x + 4, which has a non zero F intercept. But your method of finding the work done between those two values of x is correct. Alternatively you could just do the computation W = ∫Fdx, which is equivalent to your method.

In the next question, cosθ represents the cosine of the angle between the force and displacement vectors.
 
CAF123 said:
You are plotting F(x) = 2x + 4, which has a non zero F intercept. But your method of finding the work done between those two values of x is correct. Alternatively you could just do the computation W = ∫Fdx, which is equivalent to your method.
For this question, why the change in potential energy is not equal to zero?
 
haha1234 said:
For this question, why the change in potential energy is not equal to zero?
Why do you think it should be zero?
 
CAF123 said:
Why do you think it should be zero?

Since the object only moves in x direction:shy:
 
haha1234 said:
Since the object only moves in x direction:shy:
The key point here is that the force is a function of x. Since V = -W, and W was non zero, we already know that the change in potential energy is similarly non zero.

If you consider an object acted on upon by gravity moving only along the x axis. Gravity points downwards so the angle between the force vector and displacement is 90 degrees and hence W = F.dx = 0 and correspondingly the change in potential energy of the object is zero. The force in this case, though, is not a function of x.
 
CAF123 said:
The key point here is that the force is a function of x. Since V = -W, and W was non zero, we already know that the change in potential energy is similarly non zero.

If you consider an object acted on upon by gravity moving only along the x axis. Gravity points downwards so the angle between the force vector and displacement is 90 degrees and hence W = F.dx = 0 and correspondingly the change in potential energy of the object is zero. The force in this case, though, is not a function of x.

Sorry,but what is v?
And how to find the potential energy?:confused:
 
haha1234 said:
Sorry,but what is v?
V is potential energy here, perhaps you use U.

To elaborate on my example earlier, a block attached to a spring undergoes a restoring force F= -kx when it is displaced some x. We can talk about the potential energy stored in the spring even though the problem is 1 dimensional.
 
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CAF123 said:
You are plotting F(x) = 2x + 4, which has a non zero F intercept. But your method of finding the work done between those two values of x is correct. Alternatively you could just do the computation W = ∫Fdx, which is equivalent to your method.

In the next question, cosθ represents the cosine of the angle between the force and displacement vectors.

How to find the cosθ?
Is the direction of F show as follow?
And what is the direction of the displacement vector?
 

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  • #10
The work of what force is required by the problem?
 
  • #11
haha1234 said:
How to find the cosθ?
cosθ is the angle between the force vector acting on the particle and and the displacement vector of the particle. The question asks about one particular force and so you have to first identify it's direction. It is not as you have shown it.
And what is the direction of the displacement vector?
Consider first the path OAC. This is made of two segments OA and AC. The particle travels from O to C. Consider each segment separately.
 

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