Kinetic Energy Plot: Find Speed & Turning Point

[soccerkid999]

Homework Statement

Figure shows a plot of potential energy U versus position x of a 0.900 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA = 15 J, UB = 35 J and UC = 45 J. The particle is released at x = 4.5 m with an initial speed of 6.80 m/s, headed in the negative x direction.


If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point?


http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c08/q39.jpg

Homework Equations

Ek=1/2mv^2

The Attempt at a Solution

I have no clue what to do.

 

[IniquiTrance]

What is the particle's total energy at t = 0?

 

[soccerkid999]

What is the particle's total energy at t = 0?

they don't give time, help anyone?

 

[IniquiTrance]

Let me rephrase, what is the particle's total energy?

 

[rl.bhat]

Find the KE at 4 m.
UB - UA = KE at 4m - KE at 2m
If it is positive, the particle will reach 1 m.

 

[Gold3nlily]

Find the KE at 4 m.
UB - UA = KE at 4m - KE at 2m
If it is positive, the particle will reach 1 m.

Hi, I am also having trouble with this problem. It is not homework, I am doing it out of the book.

So I am trying to find the speed at x=1m

Why would KE at 1 m not be Uc - Ub? Isn't KE the space inbetween the U line and the total energy?

Just in case last person pic doesn't work any more here is my own link from book:

http://edugen.wiley.com/edugen/courses/crs4957/halliday9118/halliday9118c08/image_n/nt0055-y.gif

And after I find the KE I use
KE = 1/2 mv^2 to find the velocity

I didn't find the right velocity when I took Ub-Ua.
I also didn't find the right velocity when I took Uc - Ub.

The answer should be : 2.1 m/s

How do I get this?

 

[Gold3nlily]

would it have been better to start a new thread for my question? This last question is over a year old...

Any advice helps. Thank you! :)