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

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 [Broken]

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

Last edited by a moderator:

Related Introductory Physics Homework Help News on Phys.org
Andrew Mason
Homework Helper
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

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
Staff Emeritus
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 Vfinal is less than Vinitial. ΔV is therefore negative, and ΔU = e ΔV is also negative. The electron rolls up electric potential hills, so Vfinal is greater than Vinitial. 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 $\Delta K = -\Delta U$.