# Graph explanation question

1. Dec 20, 2008

### transgalactic

a particle is moving along the X axes under the influence of
a conservative force with a potential energy.
by this graph :
http://img376.imageshack.us/img376/5609/82453845uw5.gif [Broken]

1.describe how the particle moves with E1 energy
where E1>U1
which comes from the left x<-a
and describe how the speed of the particle changes .

here is the solution:
http://img216.imageshack.us/img216/944/46482160yg4.gif [Broken]

can you give explanation to how to get this solution
i know that force is the derivative of potential energy

??

Last edited by a moderator: May 3, 2017
2. Dec 20, 2008

### buffordboy23

How specific does the description have to be? E1 is the total energy of the particle, which is a constant. U(x) is the potential energy of the particle at position x. The kinetic energy is the difference between the two: K = E1 - U(x).

So what happens to the particle? Is it bounded? How does its kinetic energy change?

3. Dec 20, 2008

### transgalactic

i showed the solution
its in the end
i cant understand how they came to this conclusion?

4. Dec 20, 2008

### buffordboy23

E = U + K = U + (1/2)mv^2

Solve for the velocity v. The velocity of the particle changes with position because the potential is not constant. So there is a force which acts to speed the particle up and slow it down at certain positions, but you don't need to compute the force to determine what the velocity v is. Make sense yet?

5. Dec 21, 2008

### transgalactic

i cant understand this graph
there is only one axes
and some weird letters
i cant understand whats happening there?

6. Dec 21, 2008

### transgalactic

i know that U is potential energy and E is the total
one

where is A and B
??

7. Dec 21, 2008

### nrqed

The graph shows the potential energy as a function of the position of the particle.

x=-a is the point where the graph starts to go down and x=b is the point where it becomes flat again.

8. Dec 21, 2008

### buffordboy23

Also, something is wrong with the solution. The mass, m, should be located in the denominator under the radical. Look at the resultant units with your answer and you will see that they do not match those required for a velocity.