# Is this a problem of derived equation?

1. Jun 29, 2008

### Qwer

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:

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

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

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

$$v^2 = \frac{2 k(e)^2}{m^2}$$ [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 $$v^2$$ from Eq.1 like this:

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

$$2k(e) = mv^2$$

$$v^2 = \frac{2 k(e)}{m}$$

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

$$F(e) = \frac{kq_1q_2}{r^2}$$

and I don't know what she is really looking for. Thanks in advance!

2. Jun 29, 2008

### Defennder

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. Jun 29, 2008

### Qwer

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 $$v^2$$, so could you say that Eq.4 is the correct solution? Is there also a siginificance to Eq.2?

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. Jun 29, 2008

### Defennder

This happens to me all the time.

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.

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.