Does work = neg or pos change in potential energy?

[bluesteels]

Homework Statement

How much work would it take to push two protons very
slowly from a separation of 2.00 * 10^-10 m (a typical atomic distance) to 3.00 * 10^-15 m (a typical nuclear distance)?

Relevant Equations

PE = kqq'/r
q = 1.61*10^-19
r1= 2*10^-10
r2= 3*10^-15
k = (9*10^9)

u = (9*10^9)(1.61*10^-19)^2 * (1/[3*10^-15 ]- 1/[2*10^-10])

u = 7.68*10^-14 J

but here the question. I have been taught that W= -U so shouldn't the answer be negative??
When i look up at the solution all other sources say that the W = U and therefore the answer is in postive.

 

[Orodruin]

Homework Statement:: How much work would it take to push two protons very
slowly from a separation of 2.00 * 10^-10 m (a typical atomic distance) to 3.00 * 10^-15 m (a typical nuclear distance)?
Relevant Equations:: PE = kqq'/r
q = 1.61*10^-19
r1= 2*10^-10
r2= 3*10^-15
k = (9*10^9)

I have been taught that W= -U so shouldn't the answer be negative??
When i look up at the solution all other sources say that the W = U and therefore the answer is in postive.

Neither of these makes sense. A potential is not work, it is a potential. The work done by the related force depends on the difference of the potentials at the start and end points. Also note that you are asked to find the work you need to do to push the protons together, not the work done by the electromagnetic force. Since the total work needs to be zero (unless you want the protons to end up moving), these two are equal in magnitude but opposite in sign.

 

[Steve4Physics]

Homework Statement:: How much work would it take to push two protons very
slowly from a separation of 2.00 * 10^-10 m (a typical atomic distance) to 3.00 * 10^-15 m (a typical nuclear distance)?
.
.
but here the question. I have been taught that W= -U so shouldn't the answer be negative??
When i look up at the solution all other sources say that the W = U and therefore the answer is in postive.

In thermodynamics books (when there is zero heat transfer) you may see ‘W=-U’ (or equivalent). It means the work done by a system equals the system's loss in internal energy.

But you will also see (in different books) ‘W=U’. Ir means the work done on a system equals the system's gain in internal energy.

Both are correct. It’s a case of what convention is being used for the meaning of W.

You are asked how much work (W) external forces do pushing 2 protons (which repel) closer together. W is positive because each external force and its displacement have the same direction.

The system’s internal (potential) energy increases, so ‘W=U’ is appropriate.

 

[vcsharp2003]

Homework Statement:: How much work would it take to push two protons very
slowly from a separation of 2.00 * 10^-10 m (a typical atomic distance) to 3.00 * 10^-15 m (a typical nuclear distance)?
Relevant Equations:: PE = kqq'/r
q = 1.61*10^-19
r1= 2*10^-10
r2= 3*10^-15
k = (9*10^9)

but here the question. I have been taught that W= -U so shouldn't the answer be negative??

The negative sign is only used if $W$ is work done by electric force. But if $W$ is work done by external applied force then there is no negative sign. In this case, the external forces are applied to the charges to bring them closer and this work will simply be change in electrical PE of the system of charges i.e. without the negative sign.

 

[bluesteels]

The negative sign is only used if $W$ is work done by electric force. But if $W$ is work done by external applied force then there is no negative sign. In this case, the external forces are applied to the charges to bring them closer and this work will simply be change in electrical PE of the system of charges i.e. without the negative sign.

how do you figure out if it is work done by external or if something is doing work. I have a hard time figuring it out

 

[PeroK]

how do you figure out if it is work done by external? I have a hard time figuring it out

If a system gains or loses mechanical energy, then an external force must be involved. If there is no external force, then the mechanical energy (sum of KE and PE) must be constant.

In this question the phrase "very slowly" is intended to indicate that KE remains negligible throughout.

 

[vcsharp2003]

how do you figure out if it is work done by external or if something is doing work. I have a hard time figuring it out

The question is asking "How much work would it take to push two protons..." means some force is pushing the protons closer and since protons will always repel so one would have to apply a force to bring them closer. This applied force is not the electrical force and as a matter of fact any time there is an "applied" force it means it's an external force. Electrical force on the other hand is an internal force between protons since it's not an applied force. So, just remember the word applied makes it an external force. Or you could say an external force involves deliberation whereas internal forces are automatically there and require no deliberation.

 

[vcsharp2003]

The system’s internal (potential) energy increases, so ‘W=U’ is appropriate.

Shouldn't the right hand side be change in potential energy rather than just potential energy i.e. $W = \Delta U$ rather than $W=U$?

 

[Steve4Physics]

Shouldn't the right hand side be change in potential energy rather than just potential energy i.e. $W = \Delta U$ rather than $W=U$?

Yes, that is better and is the usual format.

In Post #3 I wanted to answer in terms of the OP’s terminology. That’s why I said " ‘W=-U’ (or equivalent)". By 'equivalent' I was referring to formulations using ‘Δ’.

I have also seen ΔW and ΔQ used (e.g. ΔU = ΔQ + ΔW) , but I don’t like using ΔW and am in two minds about ΔQ.

 

[vcsharp2003]

I have also seen ΔW and ΔQ used (e.g. ΔU = ΔQ + ΔW) , but I don’t like using ΔW and am in two minds about ΔQ.

Yes, I agree with you since heat energy flow Q and work done W are always absolute quantities unlike potential energy. So, using just Q and W makes sense. Potential energy is a relative quantity so change in PE i.e. $\Delta U$ makes sense.

 

[vcsharp2003]

Homework Statement:: How much work would it take to push two protons very
slowly from a separation of 2.00 * 10^-10 m (a typical atomic distance) to 3.00 * 10^-15 m (a typical nuclear distance)?
Relevant Equations:: PE = kqq'/r
q = 1.61*10^-19
r1= 2*10^-10
r2= 3*10^-15
k = (9*10^9)

I have been taught that W= -U so shouldn't the answer be negative??

Instead of the work to potential energy equation you're looking at, you could consider the following alternate equivalent equations which makes things more clear. $W_{ex}$ is work done by external force(s), whereas $W_{el}$ is work done by electrical forces.

$$W_{ex} = \Delta U$$
$$W_{el} =- \Delta U$$

 

[vcsharp2003]

how do you figure out if it is work done by external or if something is doing work. I have a hard time figuring it out

A good way of looking at external vs internal forces is to ask yourself whether the force you're looking at is between parts of the system; if yes, then it's an internal force else it's an external force. In this question, the "system" consists of the two protons. So, the electrical force between the protons is an internal force since it's a force between parts of the system we have defined. The push force, on the other hand, is an external force since it's not between the parts of our system.

 

[kuruman]

but here the question. I have been taught that W= -U so shouldn't the answer be negative??
When i look up at the solution all other sources say that the W = U and therefore the answer is in postive.

You should also have been taught that $W=\vec F\cdot \vec d$. If the external force $\vec F$ that you apply and the displacement $\vec d$ of the charges are in the same direction (as is the case here), the dot product and hence the work is positive. Note that this follows strictly from the definition of work without reference to potential energy.