Solved: Help With Energy Problem Homework

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The discussion revolves around solving a physics homework problem involving a 20kg object influenced by a conservative force. For part (a), the potential energy at x=2.0m can be calculated by integrating the force function, yielding a result of -19.3J. In part (b), the conservation of energy principle is emphasized, where the total energy (kinetic plus potential) remains constant, allowing for the calculation of speed at different positions. The participants clarify that to find the speed at the origin, one must set up an equation using energy conservation between two points. Overall, the focus is on applying energy conservation and integrating force to solve for potential energy and speed.
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HELP!Energy problem

Homework Statement


A 20kg object is acted on by a conservative force given by F=-3.0x-5.0x^2, with F in Newtons and x in meters. take the potential energy associated with the force to zero when the object is at x=0
a) what is the potential energy of the system associated with the force when the object is at x=2.0m?
b)If the object has a velocity of 4.0m/s in the negative direction of the x-axis when it is at x=5.0m, what is its speed when it passes through the origin?
c)What are the answers to a) and b) if the potential energy of the system is taken to be -8.0J when the object is at x=0.

The Attempt at a Solution


a)I can get the equation for finding out PE by integrating F=-3.0x-5.0x^2, which is -1.5x^2-5/3x^3
sub 2.0 into x
then the answer I got is -19.3
b) I don't really know how to approach this question, but I know that I have to somehow relate the conservative energy to speed...maybe to kinetic energy?1/2mv^2?
c)
a. -8=-1.5x^2-5/3x^3
x=-2.04m
but the question says when x=0, so I don't know what did i do wrong with this question
b. I didn't know how to do b), so I don't really know how to det the answer for this too.

I need some help on this. Thank you very much.
 
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(b) is simple, just put in x=2 into your answer of part (a).
 


Irid said:
(b) is simple, just put in x=2 into your answer of part (a).

Then isn't the answer same as a)
 


Oh yeah, sorry :). For part (b) you have to use energy conservation. Kinetic energy (T=mvv/2) plus potential energy is a constant at any x, so just make an equation at for two points.
 


Irid said:
Oh yeah, sorry :). For part (b) you have to use energy conservation. Kinetic energy (T=mvv/2) plus potential energy is a constant at any x, so just make an equation at for two points.


E=K+PE
find PE by plugging in the x value
and plug in the x values and m into K then I got the energy...
sorry, but how can you get the speed from the answer I got for the energy
And also do you know whether i did part c right or not?
 


Consider points x_1=0 and x_2=5.
Then by energy conservation:

\frac{mv_1^2}{2} + V(x_1) = \frac{mv_2^2}{2} + V(x_2)

You have one of the speeds, so just solve for the other one.
 
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