Potential Energy of a Particle

In summary: Ah, I see... I missed the part about the force being conservative. If a force is conservative, then W = ∆KE = -∆U, since E=KE+U and E doesn't change at all, right? so would it not be easier to simply say
  • #1
Yuravian
5
0

Homework Statement


A 0.40-kg particle moves under the influence of a single conservative force. At point A where the particle has a speed of 10 m/s, the potential energy associated with the conservative force is +40 J. As the particle moves from A to B, the force does +25 J of work on the particle. What is the value of the potential energy at point B?

The problem, for me is that there is no diagram included on the homework sent out in the email from the professor and I think the proper problem would have had a diagram of the path. I may be wrong though.

Homework Equations


Well, this is clearly a work and kinetic energy problem, so:
[tex]W=\Delta U+\Delta K[/tex]
[tex]W=Fdcos\theta[/tex] (not sure if it applies here)
[tex]\Delta K=1/2mv^{2}_{o}-1/2mv^{2}_{f}[/tex]
[tex]\Delta U=mgh_{o}-mgh_{f}[/tex] (again, I don't think this applies)

The Attempt at a Solution


straighten out my variables:
[tex]v_o=10 m/s[/tex]
[tex]K_o=1/2mv^2_o=20 J[/tex]
[tex]U_o=40 J[/tex]
[tex]W=25 J=U_f-U_o+K_f-K_o[/tex]

So at this point, I do not have the acceleration of the particle (cannot find F or v final), and no information on the final state of the particle other than the amount of work done on it. Where should I go from here?
 
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  • #2
Actually I don't think you need a diagram to solve this. Think about: if we assume that the energy is conserved in the system then the total energy is equal to the kinetic + potential energy. We know the potential energy at point a and can calculate the kinetic energy at point a as well...

It might be helpful to know that

[tex]W = \Delta K[/tex]
 
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  • #3
Sorry, I just cut that out. I mis-read the question. I still do not think it can be solved though.
 
  • #4
qspeechc said:
Sorry, I just cut that out. I mis-read the question. I still do not think it can be solved though.

No, I really think it can be solved. You can find the total mechanical energy, and then find the kinetic energy at point b and from that find the potential at point b.
 
  • #5
How would you find the kinetic energy at B?
 
  • #6
At point A: [tex]KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.40kg)(10\frac{m}{s})^2 = 20J[/tex]

[tex]E_{mec} = KE + U = 20J + 40J = 60J[/tex]

We know the work done from point A to B and
[tex]W = \Delta K = \frac{1}{2}m(v_f^2-v_o^2)[/tex]

We need to solve for the velocity of the particle at point B... we get:

[tex]v_{f}^2 = \sqrt{\frac{2W}{m}+v_o^2} = \sqrt{\frac{2(25J)}{0.40kg}+10\frac{m}{s}^2} = 15\frac{m}{s}[/tex]

So KE at B: [tex]\frac{1}{2}mv^2 = \frac{1}{2}(0.40kg)(15\frac{m}{s})^2 = 45J[/tex]

Then find the potential energy

[tex]U = E_{mec} - KE = 60J - 45J = 15J[/tex]
 
  • #7
As Yuravian states, work is change in mechanical energy, that is
W = ∆KE + ∆U

Not:
W = ∆KE
as you have stated.
 
  • #8
qspeechc said:
As Yuravian states, work is change in mechanical energy, that is
W = ∆KE + ∆U

Not:
W = ∆KE
as you have stated.

We're talking about a single conservative force acting on the particle. Conservative forces won't change the total amount of mechanical energy we have.

∆KE + ∆U = ∆E = W (by non conservative forces)
 
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  • #9
So? Can a single conservative force not change the potential energy? What if I lift a bag of sand vertically in a vacuum?
 
  • #10
qspeechc said:
So? Can a single conservative force not change the potential energy? What if I lift a bag of sand vertically in a vacuum?

It will change the potential energy, but it will not change the total amount of energy in the system.
 
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  • #11
If I do work in lifting the bag of sand, initially at rest on the floor, can I not lift it vertically some distance AND give it some speed, so that now it has BOTH potential AND kinetic energy, where originally it had none?
 
  • #12
qspeechc said:
If I do work in lifting the bag of sand, initially at rest on the floor, can I not lift it vertically some distance AND give it some speed, so that now it has BOTH potential AND kinetic energy, where originally it had none?

Sure, but ∆KE = -∆U = W, U is going to be changing negatively since the work is in the opposite direction of the force of gravity. The total energy will still be 0.
 
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  • #13
Feldoh said:
At point A: [tex]KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.40kg)(10\frac{m}{s})^2 = 20J[/tex]

[tex]E_{mec} = KE + U = 20J + 40J = 60J[/tex]

We know the work done from point A to B and
[tex]W = \Delta K = \frac{1}{2}m(v_f^2-v_o^2)[/tex]

We need to solve for the velocity of the particle at point B... we get:

[tex]v_{f}^2 = \sqrt{\frac{2W}{m}+v_o^2} = \sqrt{\frac{2(25J)}{0.40kg}+10\frac{m}{s}^2} = 15\frac{m}{s}[/tex]

So KE at B: [tex]\frac{1}{2}mv^2 = \frac{1}{2}(0.40kg)(15\frac{m}{s})^2 = 45J[/tex]

Then find the potential energy

[tex]U = E_{mec} - KE = 60J - 45J = 15J[/tex]

Ah, I see... I missed the part about the force being conservative. If a force is conservative, then W = ∆KE = -∆U, since E=KE+U and E doesn't change at all, right? so would it not be easier to simply say [tex] U_f = U_o+\Delta U = 40J - 25J = 15 J[/tex]?
 
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  • #14
Feldoh said:
Sure, but ∆KE = -∆U = W, U is going to be changing negatively since the work is in the opposite direction of the force of gravity. The total energy will still be 0.

Wait a minute. If your v is high and stays constant, wouldn't your U increase while K stayed constant? Correct me if I'm wrong, but if you lift a bag of sand, the force you put on it isn't conservative in the first place.
 
  • #15
Yuravian said:
Wait a minute. If your v is high and stays constant, wouldn't your U increase while K stayed constant? Correct me if I'm wrong, but if you lift a bag of sand, the force you put on it isn't conservative in the first place.

Oh I think he meant like throwing it upwards. I think you're right though the force of the push is not conservative...

Yuravian said:
Ah, I see... I missed the part about the force being conservative. If a force is conservative, then W = ∆KE = -∆U, since E=KE+U and E doesn't change at all, right? so would it not be easier to simply say [tex] U_f = U_o+\Delta U = 40J - 25J = 15 J[/tex]?

Haha, yeah your way sure beats mine :P
 

1. What is potential energy?

Potential energy is the energy that an object possesses due to its position or configuration in a system. It is stored energy that can be released and converted into other forms, such as kinetic energy.

2. What is the formula for calculating potential energy of a particle?

The formula for potential energy of a particle is PE = mgh, where m is the mass of the particle, g is the acceleration due to gravity, and h is the height of the particle above a reference point.

3. How does potential energy change with position?

Potential energy changes with position because it is dependent on the gravitational force acting on the particle. As the particle moves closer to the reference point, its potential energy decreases, and as it moves further away, its potential energy increases.

4. What are some examples of potential energy in everyday life?

Some examples of potential energy in everyday life include a rollercoaster at the top of a hill, a stretched rubber band, a compressed spring, and a book sitting on a shelf.

5. Can potential energy be negative?

Yes, potential energy can be negative if the reference point is chosen to be at a higher position than the particle. This indicates that the particle has less energy compared to when it is at the reference point, and the potential energy is negative in value.

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