# Energy conservation problem

1. Oct 26, 2008

### Benzoate

1. The problem statement, all variables and given/known data
A particle P of mass m , which is on the negative x-axis. is moving towards the origin with a constant speed u. When P reaches the origin , it experiences the force F=-Kx^2, where K is a positive constant. How far does P get along the positive x-axis?

2. Relevant equations

.5*m*v^2+V(x)=E(x)

3. The attempt at a solution

dx/dt=[2*(E-V(x))]^.5

dV/dx=-F=Kx^2 ==> V=Kx^3/3

T0+V0=T1+V1

1/2*m*v^2+Kx^3/3=.5*m*u^2+0 since as the speed of particle increases V is zero. The problem with that explanation is, the speed is constant so the speed doesn't increase or decrease.

2. Oct 26, 2008

### Hootenanny

Staff Emeritus
The speed is constant until the particle reaches the origin, after which it accelerates (since there is a force acting on it).

3. Oct 28, 2008

### Benzoate

But my final KE is still a non-zero number and my final V is still zero right? Show I look at only the x-intervals when the particle passes the origin therefore setting my initial x-value to be at zero?

4. Oct 28, 2008

### Hootenanny

Staff Emeritus
I don't understand this statement. If your final velocity is zero (as required by the question) then the final kinetic energy must be zero, be definition.

5. Oct 28, 2008

### Benzoate

But if my particle is going to accelerate once it passes the origin, then that means the velocity of particle increases and therefore my potential energy is zero.

6. Oct 28, 2008

### Hootenanny

Staff Emeritus
An acceleration is simply a change in velocity, whether that change be an increase or decrease.

7. Oct 28, 2008

### Benzoate

Should I count the time the speed of the particle is constant on the x-axis or should I just count the time the particle is only accelerating?

Otherwise I would used this equation to find out far the particle has travel:

dx/dt=[2*(E-V(x))]^.5

Would I used the conservation of energy to find the final kinetic energy which would be needed to find the total energy?

My only concern now is how would I calculate the time since I need to integrate dx/dt with respect to time to obtain the total position.

8. Oct 28, 2008

### Hootenanny

Staff Emeritus
There is no need for any integration here. Simply write down an expression for conservation of energy.

9. Oct 28, 2008

### Benzoate

Okay. T+V=E

T0+V0=T1+V1

T0=mu^2/2

V0=0

T1=m*v^2/2

V1=Kx^3/3

10. Oct 28, 2008

### Hootenanny

Staff Emeritus
Good. What can you say about T1 when the particle is at it's maximum displacement?

11. Oct 28, 2008

### Benzoate

at maximum displacement, T1=0

12. Oct 29, 2008

### Hootenanny

Staff Emeritus
Correct. So can you now write down an equation relatics the initial kinetc energy and the final potential energy of the particle?

13. Oct 29, 2008

yes. Thanks.