Solving Energy of a System - Plotting Graphs & Calculating Cosθ

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The discussion focuses on solving energy-related problems involving work and potential energy. Participants analyze the calculation of work done using the force function F(x) = 2x + 4 and confirm that the method of integrating W = ∫Fdx is valid. The concept of cosθ is clarified as the angle between the force and displacement vectors, emphasizing the importance of identifying their directions. A key point raised is that the change in potential energy is non-zero due to the force being a function of x, contrasting with scenarios where forces are constant. The conversation underscores the relationship between work, potential energy, and the angles involved in these calculations.
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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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