Why is my book using different equations to solve for velocity?

[outxbreak]

I am having a difficult time trying to figure out why my book says that I should solve for v by:
(1) Set the kinetic energy equal to the change in potential energy and solve for v

Then 3 problems later it says solve for v by:
(2)Set ΔK = −ΔU and solve for v

http://img593.imageshack.us/img593/1995/1idz.jpg


I just need to know why they are using different equations to solve for v.. this is not making any sense.

 

[Andrew Mason]

I am having a difficult time trying to figure out why my book says that I should solve for v by:
(1) Set the kinetic energy equal to the change in potential energy and solve for v

Then 3 problems later it says solve for v by:
(2)Set ΔK = −ΔU and solve for v


I just need to know why they are using different equations to solve for v.. this is not making any sense.

They are essentially the same equations. ΔK = −ΔU means that the gain in kinetic energy is equal in magnitude to the decrease in potential energy.

The V (upper case) in problem 21 is the potential difference. The v (lower case) is the velocity of the charged particle.

AM

 

[outxbreak]

They are essentially the same equations. ΔK = −ΔU means that the gain in kinetic energy is equal in magnitude to the decrease in potential energy.

But how do I know when to use
ΔK = −ΔU
or
ΔK = ΔU

That is my problem. :'(

 

[vela]

The book's being sloppy in the first case. It's really using $\Delta K = \lvert \Delta U \rvert$. For both the electron and proton, the potential energy decreases, so $\Delta U < 0$.

The proton rolls down electric potential hills, so V_final is less than V_initial. ΔV is therefore negative, and ΔU = e ΔV is also negative. The electron rolls up electric potential hills, so V_final is greater than V_initial. ΔV is therefore positive. The electron has a negative charges, so ΔU = -e ΔV again is negative. Because ΔU is negative, if you try to use $\frac{1}{2}mv^2 = \Delta U$, there'd be no real solution for v. You need to use $\Delta K = \frac{1}{2}mv^2 = -\Delta U$. In other words, you always use $\Delta K = -\Delta U$.

In electromagnetics, it's pretty easy to make sign mistakes if you try to rely solely on the math. It's often easier simply to neglect the signs and calculate the magnitude of quantities and then insert the correct sign based on physical intuition. For instance, in both cases here, we know that the particles are going to speed up, so their kinetic energy increases. This increase comes at the expense of a decrease in potential energy. To find the speed, we don't really care about the sign of ΔU; we just need its magnitude.

 

[Nickie]

There could be a few reasons why your book is using different equations to solve for velocity. One possibility is that the book is presenting different methods or approaches to solving for velocity, which can be helpful for understanding the concept from different perspectives. Another reason could be that the equations are being applied to different scenarios or problems, which may require different equations to accurately solve for velocity. It is also possible that the equations are simply different forms of the same equation, just rearranged in different ways for convenience or clarity. Ultimately, the important thing is to understand the underlying principles and concepts behind the equations, rather than just memorizing specific equations. This will allow you to apply the appropriate equation for any given situation and solve for velocity accurately. If you are still having trouble understanding why different equations are being used, it may be helpful to consult with your teacher or a tutor for further clarification.