Physics work question

[matt102498]

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I'm studying using the NOVA physics book now from front to back because I found the EK book to be a little weak in certain areas though very helpful overall. I don't think the NOVA book would have been enough on it's own either though.

Anyway, my question is in the nova book work is defined W=FdeltaX, whereas in the EK book it says W=Fd and is independent of displacement. I'm tring to read ahead in NOVA but this contradiction in my head is slowing me down. Can someone here clarify? Thanks.

 

[Ontology]

It's the same thing because delta x is the change in displacement which would equal d. So work is the force parallel multiplied by displacement which can be written as d or the change in x (delta x).

 

[MedPR]

I'm studying using the NOVA physics book now from front to back because I found the EK book to be a little weak in certain areas though very helpful overall. I don't think the NOVA book would have been enough on it's own either though.

Anyway, my question is in the nova book work is defined W=FdeltaX, whereas in the EK book it says W=Fd and is independent of displacement. I'm tring to read ahead in NOVA but this contradiction in my head is slowing me down. Can someone here clarify? Thanks.

Did you mean to say W=Fd and is dependent of displacement? If there's no displacement, there is no work.

 

[ljc]

Did you mean to say W=Fd and is dependent of displacement? If there's no displacement, there is no work.

Independent is correct here, as it is the total distance traveled that determines the work done. For instance, if a ball is pushed in a circle, displacement would be zero, but that does not mean work is zero. So in W=fd, d is distance traveled, not displacement. Unless you are talking about net work, in which case d would be displacement.

If the object travels in a straight line, the two equations are the same.

 

[MedPR]

Independent is correct here, as it is the total distance traveled that determines the work done. For instance, if a ball is pushed in a circle, displacement would be zero, but that does not mean work is zero. So in W=fd, d is distance traveled, not displacement. Unless you are talking about net work, in which case d would be displacement.

If the object travels in a straight line, the two equations are the same.

I thought vector * scalar was always a vector? Work is a scalar, so doesn't W=Fd need to be displacement (vector) and not distance (scalar)?

 

[MT Headed]

MedPR is correct. The d is a displacement vector. And if you push a ball in a horizontal or vertical or any other kind of circle, returning it to its original position and at rest, the amount of work done is zero.

 

[ljc]

Whoops. My mistake. Probably better to think of work as change in energy. In my example, an object pushed in a circle will have no more energy (kinetic or potential) when it returns to d=0 than it did when it started.

 

[MedPR]

Whoops. My mistake. Probably better to think of work as change in energy. In my example, an object pushed in a circle will have no more energy (kinetic or potential) when it returns to d=0 than it did when it started.

No it wont. Work = deltaKE. If work = 0, then change in energy = 0. KEf = KEi.

It doesn't matter how you "think of work." No displacement = no work, period.

 

[MT Headed]

We need to be VERY careful with our terms here. If I lift a box from the floor and set it down on a table, I have applied a force in the direction of a displacement. I have done positive work on the box. But the kinetic energy is the same before and after the action (zero).

There was a stupid and frustrating standalone question in one of the later AAMC's (9 10 or 11) that I got wrong that thinly sliced the exact definitions of these terms. I'm still convinced somedays that I was right and they were wrong.

 

[MedPR]

We need to be VERY careful with our terms here. If I lift a box from the floor and set it down on a table, I have applied a force in the direction of a displacement. I have done positive work on the box. But the kinetic energy is the same before and after the action (zero).

There was a stupid and frustrating standalone question in one of the later AAMC's (9 10 or 11) that I got wrong that thinly sliced the exact definitions of these terms. I'm still convinced somedays that I was right and they were wrong.

In this case the work done was used to create potential energy right?

 

[MT Headed]

Right. The first law of thermodynamics says that the work you add to a system changes the total (kinetic+potential+elastic+etc) energy of the system.

But the work energy theorem says that the total work done by the net forces acting on a system generate its change in kinetic energy.

We can resolve this seeming paradox in my box example by noting that gravity is doing negative work on the box at the same time. After all, the force of gravity (down) is opposite the displacement (up).

You would think such fine distinctions wouldn't matter for the mcat, but then an aamc question pops up and I get a 14 on the PS of my practice test 🙁 I tend to remember those.

 

[ljc]

No it wont. Work = deltaKE. If work = 0, then change in energy = 0. KEf = KEi.

It doesn't matter how you "think of work." No displacement = no work, period.

You just said exactly what I said? I was agreeing with you.

 

[MedPR]

You just said exactly what I said? I was agreeing with you.

my fault, I read "will have more kinetic energy" not "will have no more kinetic energy"

 

[matt102498]

I never stopped by to say thanks, but thanks! Seems to be a tough topic for most. Glad I'm not alone.

 

[silverice]

Right. The first law of thermodynamics says that the work you add to a system changes the total (kinetic+potential+elastic+etc) energy of the system.

But the work energy theorem says that the total work done by the net forces acting on a system generate its change in kinetic energy.

We can resolve this seeming paradox in my box example by noting that gravity is doing negative work on the box at the same time. After all, the force of gravity (down) is opposite the displacement (up).

You would think such fine distinctions wouldn't matter for the mcat, but then an aamc question pops up and I get a 14 on the PS of my practice test 🙁 I tend to remember those.

So is there work or no work? Maybe I should ask do we go with the work energy theorem or first law of thermodynamic on this one?

 

[milski]

So is there work or no work? Maybe I should ask do we go with the work energy theorem or first law of thermodynamic on this one?

The total work done on the object is 0. The work done on the object by the force lifting it is X, the work done on the object by the gravity force is -X.

Using precise terminology tends to help - you should talk either about "work done on an object by a force" or "total work". Skipping the object or the force in the phrase when not talking about total work tends to make things ambiguous.