Is this a problem of derived equation?

  • Thread starter Qwer
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In summary, the conversation revolves around the four equations for kinetic energy and the confusion surrounding their significance and how they relate to a problem involving Coulombic force. The first equation, k(e) = (mv^2)/2, is the formula for kinetic energy, while the rest of the equations are just different ways of solving for v^2. It is not clear where Eq.2 comes from and there is no information about the problem that the teacher is expecting the students to solve.
  • #1
Qwer
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We had this at our physics class, but since my concern only involves computation, I posted it here. I admit I was not paying attention and was just jotting down what my teacher was scribbling on the board. These are the equations in the exact order she wrote it:

[tex]k(e) = \frac{mv^2}{2}[/tex] [Eq.1]

[tex]v = \frac{2 k(e)}{m}[/tex] [Eq.2]

[tex]\frac{mv^2}{m} = \frac{2 k(e)}{m}[/tex] [Eq.3]

[tex]v^2 = \frac{2 k(e)^2}{m^2}[/tex] [Eq.4]


I.. quite.. do not follow the significance of Eq.2. And I'm not sure my teacher wrote Eq.4 correctly, because I would simply get [tex]v^2[/tex] from Eq.1 like this:

[tex]k(e) = \frac{mv^2}{2}[/tex] [Eq.1]

[tex]2k(e) = mv^2[/tex]

[tex]v^2 = \frac{2 k(e)}{m}[/tex]


So, I'm hoping someone could explain whether Eq.2 is really a derived equation.. and if it is, to be patient enough to explain how my teacher got from Eq.1 to Eq.4. I don't really know what's going on, and she's requiring us to solve the equation

[tex]F(e) = \frac{kq_1q_2}{r^2}[/tex]

and I don't know what she is really looking for. Thanks in advance!
 
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  • #2
I can't really tell what you're writing here. The first expression is that of kinetic energy 1/2 mv^2. And the rest of the four equations seem to be just solving for v^2. And more importantly what do you mean by "solve the equation"? What exactly do you wan to solve it for? The final expression is just the Coulombic force exerted by 2 point charges on each other. I don't see what has that got to do with KE.
 
  • #3
Defennder said:
I can't really tell what you're writing here. The first expression is that of kinetic energy 1/2 mv^2. And the rest of the four equations seem to be just solving for v^2.

Thanks Defennder.

I was not really paying attention while copying from the board. Thanks for pointing out that it's really a kinetic energy formula. I understand now what my teacher meant by "k(e)". True, it is about solving for [tex]v^2[/tex], so could you say that Eq.4 is the correct solution? Is there also a siginificance to Eq.2?


And more importantly what do you mean by "solve the equation"? What exactly do you wan to solve it for? The final expression is just the Coulombic force exerted by 2 point charges on each other. I don't see what has that got to do with KE.

Thanks again for pointing out what the "F(e)" formula really is. We're still starting physics, so I'm sorry if I'm sounding clueless on physics concepts. Again, my concern is pure computation. You're right, "F(e)" does not have anything to do with KE. It is another problem that she is expecting us to "solve", based on the computation process she made on the KE formula.

Thanks Defennder for bearing with my concerns. I hope you (and others) could again help with any of the questions above.
 
  • #4
Qwer said:
I was not really paying attention while copying from the board.
This happens to me all the time.

I understand now what my teacher meant by "k(e)". True, it is about solving for [tex]v^2[/tex], so could you say that Eq.4 is the correct solution? Is there also a siginificance to Eq.2?
I don't know where Eq 2 comes from. It looks as though it should the RHS should be under a square root, if you express KE = 1/2 mv^2 in terms of v. As for the physical significance, it really depends on what problem you are solving. You may or may not be required to find v, given the other variables.

It is another problem that she is expecting us to "solve", based on the computation process she made on the KE formula.
I believe the problem you posted is incomplete. So I really don't know how to help here. There are many ways in which you need to use the kinetic energy of something to find the work done by a force (something known as the work-energy theorem which you may or may not have learnt) That's about the only thing I can think of where you'll actually need to use the KE formula when solving for force. But then again there are many other possible physical possibilities such as that force formula you saw on the board may actually be required for a different part of the same question. So I really can't tell how they are related unless you post the question she was solving.
 

1. What is a derived equation?

A derived equation is an equation that is obtained from other equations through mathematical operations. It is often used to simplify complex equations or to find a relationship between different variables.

2. How do I know if a problem involves derived equations?

If the problem involves finding a relationship between variables or simplifying equations, it is likely that derived equations will be involved. Look for key words such as "derive", "simplify", or "find the equation" in the problem statement.

3. What is the purpose of using derived equations in scientific research?

Derived equations allow scientists to simplify complex relationships between variables and make predictions or draw conclusions from the data. They also help to identify patterns and relationships that may not be immediately obvious.

4. Can derived equations be used in all fields of science?

Yes, derived equations are used in various fields of science such as physics, chemistry, biology, and engineering. They are a fundamental part of scientific research and are used to understand and analyze complex systems and phenomena.

5. How do I create a derived equation?

To create a derived equation, you need to use mathematical operations such as addition, subtraction, multiplication, and division on existing equations. It is important to follow the rules of algebra and to simplify the equation as much as possible to obtain a clear relationship between the variables.

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