Work-Energy Theorem: Understanding Vertical Movement

[Physicsnuubie]

Hi, i am having a slight confusion with this theorem.

I understand that if a car travels horizontally for s m at the uniform acceleration, the
Net Work Done = Change in K.E. (by Work-Energy Theorem)
The change in K.E. is the amount of joules required to exert the amount of Net Force on the car to move a distance of s m.

However, if the car is moving diagonally 45 degress upwards, then
Net Work Done = Change in K.E. + Change in G.P.E. (by manual logical method to find work done)
so, the total Change in K.E. + Change in G.P.E. is the amount of joules required to exert the amount of Net Force on the car to move 45 degrees diagonally upwards.

I am just wondering, is it possible to apply Work-Energy Theorem in cases where a body is involved in vertical movement... Because, in the second scenario above, i don't see any work-energy theorem being applied. Furthermore, Work-Energy Theorem only states that Net Work Done = Change in K.E. ;no other kinds of energy was included in the formula,

Can someone please enlighten me? Very much appreciated. Thank You!

 

[Doc Al]

The Work-KE theorem states that the net work done by all forces equals the change in KE. But instead of explicitly treating gravity as a force, we often include it as an energy term--gravitational PE. The change in PE represents the work done by gravity. (But you can always use the general form of the law as long as you include the work done by all forces.)

So:
(1) Net Work Done (by all forces, including gravity) = Change in K.E.
(2) Net Work Done (by all forces, except gravity) = Change in K.E. + Change in G.P.E.

These are equivalent statements. Your choice as to which you use.

 

[Physicsnuubie]

Hi,
thank you very much for enlightening me.

If I did not mistaken you, what you mean is,
Work-Energy Theorem
Net Work Done = Change in K.E., where change in K.E. = Change in G.P.E. + Change in K.E₁.
so the problem here lies in the term(change in K.E.) being used. correct?
if the Net Work Done in Work-Energy Theorem includes other forms of energy(GPE), isn't it confusing call BOTH forms of energies as just K.E. ??


In conclusion, it seems like the term used in Work-Energy Theorem is confusing...

 

[Doc Al]

If I did not mistaken you, what you mean is,
Work-Energy Theorem
Net Work Done = Change in K.E., where change in K.E. = Change in G.P.E. + Change in K.E₁.

No. How can change in KE = change in KE + something else ??
(I meant what I said.)

so the problem here lies in the term(change in K.E.) being used. correct?
if the Net Work Done in Work-Energy Theorem includes other forms of energy(GPE), isn't it confusing call BOTH forms of energies as just K.E. ??

It would be confusing, but that's not what I did. KE and PE are different things.

Please reread my last post. Pay particular attention to what is included in the "Work Done" term.

 

[Physicsnuubie]

(by all forces, including gravity)

alright, does your including gravity mean the weight of the car? or do u mean the sum of upward force & the weight of the car?


because,
now that the car is moving 45 degrees upwards at a constant acceleration.
to find the total work done from point A(bottom of a slope) at rest, to point B(top of slope) which is 10meter above point A, how can we use Work Energy Theorem to find the total work done?

Supposedly we use Work Energy Theorem,
Net Work Done = 0.5(m_{of car})(v²_{pt B}) - 0

In this case, what does the answer 0.5(m_{of car})(v²_{pt B}) represents?

Option 1) does the answer represents K.E. + P.E. ?
or
Option 2) only K.E.


when the car moves from point A to B, certainly, a GPE is incurred. but where is it written in the workings?

 

[Doc Al]

alright, does your including gravity mean the weight of the car? or do u mean the sum of upward force & the weight of the car?

It means the work done by all forces, including gravity (the weight of the car).

because,
now that the car is moving 45 degrees upwards at a constant acceleration.
to find the total work done from point A(bottom of a slope) at rest, to point B(top of slope) which is 10meter above point A, how can we use Work Energy Theorem to find the total work done?

Supposedly we use Work Energy Theorem,
Net Work Done = 0.5(m_{of car})(v²_{pt B}) - 0

In this case, what does the answer 0.5(m_{of car})(v²_{pt B}) represents?

Option 1) does the answer represents K.E. + P.E. ?
or
Option 2) only K.E.

Change in KE always means change in KE, of course.

when the car moves from point A to B, certainly, a GPE is incurred. but where is it written in the workings?

If you want to use GPE to represent gravity, then this equation applies:
Net Work Done (by all forces, except gravity) = Change in K.E. + Change in G.P.E.
or
Change in K.E. = Net Work Done (by all forces, except gravity) - Change in G.P.E.

If you didn't want to use GPE, then this version applies:
Net Work Done (by all forces, including gravity) = Change in K.E.
or
Change in K.E. = Net Work Done (by all forces, including gravity)

 

[Physicsnuubie]

What my solution to my question is:

Resolving the horizontal component of velocity at point B, then using K.E. formula, 0.5(m)(v²) to find the K.E. gained.

Next step,
Finding GPE, which is, GPE = (m)(9.81)(10)

Next,
Sum up K.E. and GPE., and that will be the NET work done for the car to move from point A to point B.


The problem is, I don't see any Work-Energy Theorem involved.
Rising questions:
1) Can Work-Energy Theorem, i.e. Net work done = Change in K.E. be applied here?
2) Have I applied Work-Energy Theorem unknowingly?

 

[Doc Al]

What my solution to my question is:

Resolving the horizontal component of velocity at point B, then using K.E. formula, 0.5(m)(v²) to find the K.E. gained.

Next step,
Finding GPE, which is, GPE = (m)(9.81)(10)

Next,
Sum up K.E. and GPE., and that will be the NET work done for the car to move from point A to point B.

That will be the net work done on the car by all forces except gravity. (You've already included the effect of gravity by using the GPE term.)

The problem is, I don't see any Work-Energy Theorem involved.
Rising questions:
1) Can Work-Energy Theorem, i.e. Net work done = Change in K.E. be applied here?
2) Have I applied Work-Energy Theorem unknowingly?

Yes and yes.

Let's try again:
Net Work Done = ΔKE

We can break the Net work done into two pieces:
Net Work Done = Work Done (by all forces except gravity) + Work Done by gravity

So:
Work Done (by all forces except gravity) + Work Done by gravity = ΔKE

But the work done by gravity = -mgh
So:
Work Done (by all forces except gravity) -mgh = ΔKE

Or:
Work Done (by all forces except gravity) = ΔKE + mgh = ΔKE + ΔGPE

Making any sense? (It's all the same theorem, just different ways of accounting for the effect of gravity. All get the same answer, of course.)

 

[Physicsnuubie]

Work Done (by all forces except gravity) + Work Done by gravity = ΔKE

Exactly. This is why previously, I said:

Net Work Done = Change in K.E., where change in K.E.(ΔKE) = Change in G.P.E. + Change in K.E₁.

But the work done by gravity = -mgh

So, work done by car is +mgh. yea?

Or:
Work Done (by all forces except gravity) = ΔKE + mgh = ΔKE + ΔGPE

This somehow contradicts to "Work Done (by all forces except gravity) + Work Done by gravity = ΔKE" you wrote.




In conclusion, Net Work Done = ΔKE cannot be used directly if the movement involves moving vertically. yea? Unless, the ΔKE = Work Done (by all forces except gravity) + Work Done by gravity; like what u have written. And this is why i said, the theorem is confusing by terming Work Done (by all forces except gravity) & Work Done by gravity, which are 2 different forms of energies as just ΔKE.

Did I summaries any part wrongly? Please point out. Thank you very much for your patience.

 

[Physicsnuubie]

Or:
Work Done (by all forces except gravity) = ΔKE + mgh = ΔKE + ΔGPE

This somehow contradicts to "Work Done (by all forces except gravity) + Work Done by gravity = ΔKE" you wrote.

Oops. sorry. I suppose your mgh refers to work done by gravity and not by car. yea? In that sense, there will have no contradictions. hehh. :shy:

 

[Doc Al]

Exactly. This is why previously, I said:

Net Work Done = Change in K.E., where change in K.E.(ΔKE) = Change in G.P.E. + Change in K.E₁.

I don't like that version, since you created a new kind of KE. There's only one KE, based on the actual speed of the object. I think that will just add to the confusion.

But the work done by gravity = -mgh

So, work done by car is +mgh. yea?

Yes, the work done by the car against gravity is +mgh. (But we only care about work done on the car, not by the car.)

Or:
Work Done (by all forces except gravity) = ΔKE + mgh = ΔKE + ΔGPE

This somehow contradicts to "Work Done (by all forces except gravity) + Work Done by gravity = ΔKE" you wrote.

Where's the contradiction? (It's exactly the same statement!) Recall that 'Work Done by gravity' = -mgh.

In conclusion, Net Work Done = ΔKE cannot be used directly if the movement involves moving vertically. yea? Unless, the ΔKE = Work Done (by all forces except gravity) + Work Done by gravity; like what u have written.

But since that statement is always true, you can always use it!

And this is why i said, the theorem is confusing by terming Work Done (by all forces except gravity) & Work Done by gravity, which are 2 different forms of energies as just ΔKE.

They are just two different contributions to the net work done. Each force can be thought of independently and its work calculated. Add them up to get the total work done. Alternatively, you can just find the net force and compute the work done by it directly. Same thing.

 

[Doc Al]

Oops. sorry. I suppose your mgh refers to work done by gravity and not by car. yea? In that sense, there will have no contradictions. hehh. :shy:

Cool.