What Is the Potential Energy at Point B?

[Yuravian]

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:
W=\Delta U+\Delta K
W=Fdcos\theta (not sure if it applies here)
\Delta K=1/2mv^{2}_{o}-1/2mv^{2}_{f}
\Delta U=mgh_{o}-mgh_{f} (again, I don't think this applies)

The Attempt at a Solution

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

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?

 

[Feldoh]

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

W = \Delta K

 

[qspeechc]

Sorry, I just cut that out. I mis-read the question. I still do not think it can be solved though.

 

[Feldoh]

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.

 

[qspeechc]

How would you find the kinetic energy at B?

 

[Feldoh]

At point A: KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.40kg)(10\frac{m}{s})^2 = 20J

E_{mec} = KE + U = 20J + 40J = 60J

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

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

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}

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

Then find the potential energy

U = E_{mec} - KE = 60J - 45J = 15J

 

[qspeechc]

As Yuravian states, work is change in mechanical energy, that is
W = ∆KE + ∆U

Not:
W = ∆KE
as you have stated.

 

[Feldoh]

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)

 

[qspeechc]

So? Can a single conservative force not change the potential energy? What if I lift a bag of sand vertically in a vacuum?

 

[Feldoh]

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.

 

[qspeechc]

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?

 

[Feldoh]

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.

 

[Yuravian]

At point A: KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.40kg)(10\frac{m}{s})^2 = 20J

E_{mec} = KE + U = 20J + 40J = 60J

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

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

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}

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

Then find the potential energy

U = E_{mec} - KE = 60J - 45J = 15J

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 U_f = U_o+\Delta U = 40J - 25J = 15 J?

 

[Yuravian]

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.

 

[Feldoh]

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...

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 U_f = U_o+\Delta U = 40J - 25J = 15 J?

Haha, yeah your way sure beats mine :P