Distance traveled by a particle using Potential energy function

[MissBisson]

Homework Statement

A 1kg particle is in a potential given in joules by U(x)=3x2+4x-5, where x is in metres. The particle is initially at position x=1m and moving with velocity v=4m/s in the +x direction.

Find the position when the particle first comes to rest.

Homework Equations

dKe = Kf-Ki , Kf=0
Ke=-1/2mvi^2
F=ma
w=F x d

The Attempt at a Solution

U'(x) = 6x+4
Force at 1m = 10N
Ke=-1/2(1kg)(4)^2 = -8J
-8=10xd
d=-0.8m
d=xf-xi
-0.8=xf-1
xf=0.2m
However this was wrong. i also tried using several different methods but they were all wrong ..
Help would be appreciated !

 

[haruspex]

I see no need to consider forces - just work with ... work.
What are the initial and final PEs and KEs?

 

[MissBisson]

Alright so,
Ui = 2J (at x=1) Uf=0J (at rest) dU=-2J
Ki = 8J (from 1/2mv^2) Kf=0J (at rest) dK=-8J
W= dK+dU = -2 + -8 = -10J
-dU/dx = F = -6x-4, therefore at x=1 F=-10N
W= F x d
-10J = -10N x d d=1?

 

[haruspex]

Uf=0J (at rest)

No. If it finshes at position x_f, what is U there? Remember, in this question U represents only the potential energy.

-dU/dx = F

As I said, you don't need to work with forces at all.

 

[MissBisson]

if i don't use forces, how do i find my distance ? I am confused...

 

[haruspex]

if i don't use forces, how do i find my distance ? I am confused...

The force varies. You cannot here simply multiply force by distance to get energy. It requires an integration. So you might just as well work only with energy.
You correctly calculated the initial potential energy. Suppose the final position is x_f. Using the same method, calculate the final potential energy in terms of x_f.

 

[MissBisson]

Ahhh i see! So i solved the quadratic function and it gave me a positive and negative number.. i entered the positive number however that was wrong..? is it possible for the particle to have a negative final position?

 

[haruspex]

Ahhh i see! So i solved the quadratic function and it gave me a positive and negative number.. i entered the positive number however that was wrong..?

That should have worked. Please post your working.

 

[MissBisson]

Okay, my procedure was;

U(xf) = 3(xf)^2 +4(xf) -5

xf = (-4 +/- sqrt(4^2-(4*3*-5)))/2*3
xf = (-4 +/- sqrt(16-(-0)))/6
xf = (-4 + sqrt(76))/6 = 0.786m
xf = (-4 - sqrt(76))/6 = -2.119m

 

[haruspex]

Okay, my procedure was;

U(xf) = 3(xf)^2 +4(xf) -5

xf = (-4 +/- sqrt(4^2-(4*3*-5)))/2*3
xf = (-4 +/- sqrt(16-(-0)))/6
xf = (-4 + sqrt(76))/6 = 0.786m
xf = (-4 - sqrt(76))/6 = -2.119m

You appear to have solved the equation U(x_f) = 0. That is the wrong equation. What equation do you have for U(x_f) ?

 

[MissBisson]

The is the equation i have:

U(xf) = 3(xf)^2 +4(xf) -5 , do i have to set it equal to the initial potential energy which was 2 ? or do i have to somehow include the kinetic energy in the equatin also?

 

[mfb]

You have to set it equal to the initial energy. The initial energy comes both from potential and kinetic energy.

 

[MissBisson]

Thank you so much! it finally worked :)

 

[omar moured]

is the answer ( 5/3 m ) = 1.7 approximately ?

 

[haruspex]

Thank you so much! it finally worked :)

So, do you now understand that to solve it you needed to apply conservation of work, initial (PE+KE) = final (PE+KE)?

 

[MissBisson]

Yes i do thank you very much :)

 

[MissBisson]

is the answer ( 5/3 m ) = 1.7 approximately ?

Yes it was 1.67m