How can the distance traveled by a particle experiencing a force be determined?

[Benzoate]

Homework Statement

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?

Homework Equations

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

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.

 

[Hootenanny]

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

 

[Benzoate]

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

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?

 

[Hootenanny]

But my final KE is still a non-zero number and my final V is still zero right?

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.

 

[Benzoate]

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.

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.

 

[Hootenanny]

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.

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

 

[Benzoate]

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

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.

 

[Hootenanny]

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

 

[Benzoate]

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

Okay. T+V=E

T₀+V₀=T₁+V₁

T₀=mu^2/2

V₀=0

T₁=m*v^2/2

V₁=Kx^3/3

 

[Hootenanny]

Okay. T+V=E

T₀+V₀=T₁+V₁

T₀=mu^2/2

V₀=0

T₁=m*v^2/2

V₁=Kx^3/3

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

 

[Benzoate]

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

at maximum displacement, T₁=0

 

[Hootenanny]

at maximum displacement, T₁=0

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

 

[Benzoate]

yes. Thanks.