W=KE+PE (is this correct?)

[johnwandering]

I have this written down in my notes...

However, it doesn't seem correct

I know that W=change in energy, not energy...



Also...
How are the three values (Force, Work, Energy) all related?
W is Fxdistance
W is Change in E

But how is F related to E???

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[G1SG2]

I have this written down in my notes...

However, it doesn't seem correct

I know that W=change in energy, not energy...



Also...
How are the three values (Force, Work, Energy) all related?
W is Fxdistance
W is Change in E

But how is F related to E???

Hmm, Work=FxD. Work also equals change in KE. So FxD=change in KE. Someone correct me if I'm wrong.

 

[johnwandering]

Hahahaha

I am going to fail this test so hard

 

[johnwandering]

So I assume the equation is wrong???

 

[golgiapparatus88]

If I recall, work is equal to kinetic energy and NEGATIVE potential energy.

(KE) = (-PE)

 

[tlaloc87]

I don't know why I bother answering these questions over and over...

 

[johnwandering]

What???
My question pertains to Physics


I thought W was change in E
Not E... Am I mistaken??


so W=KE=-PE??


That doesn't sound right at all...

 

[Natasha1989]

If there are no nonconservative forces at work, delta E = delta K + delta U = 0

If there are nonconservative forces, delta W = delta E = delta K + delta U

 

[tlaloc87]

I don't know why I bother answering these questions over and over...

Did you think that physics and chemistry are completely unrelated and that the explanation of work in terms of thermodynamics has nothing to do with physics in the slightest?

 

[g8orlife]

Also...
How are the three values (Force, Work, Energy) all related?
W is Fxdistance
W is Change in E

But how is F related to E???

(FYI: [d] = "delta" or "change in")

Einitial + Work = Efinal
thus (PEinitial + KEinitial) + W = (PEfinal+ KEfinal)

W = Fdcos(theta) = mad = [d]PE = mgh = 1/2k(x^2) = [d]KE = 1/2m(v^2) = Joules = [d]E = mc[d]T = fd = "mu" FNd = "mu" mgd = Newton*meter = (kg*m^2)/(s^2) = Power*time = (I^2)Rt = IVt = Watt*seconds ... etc.

 

[Frazier]

What???
My question pertains to Physics


I thought W was change in E
Not E... Am I mistaken??


so W=KE=-PE??
That doesn't sound right at all...

Work is the change in KE.
Work also equals the change in (-PE).

Why? For PE to change typically means that KE has changed (being that they are related in a sense).

Think about holding a pen at shoulder height -- it has a given PE at this location and zero KE if it isn't moving.

SUDDENLY you move the pen down to your knee area (during this period respective changes in KE and PE are simultaneously taking place). The PE has gotten smaller [and thus the change in PE is a negative value]. You take the negative of this negative value (a la "-PE") and you get a positive value (your change in KE). The pen moved a certain distance from your shoulder to your knee so thus the value for the (change in KE) = W = (change in -PE).

 

[lilqu]

do you have a specific MCAT question youre struggling with? youll be able to solidify these consepts.

 

[g8orlife]

I don't know why I bother answering these questions over and over...

Did you think that physics and chemistry are completely unrelated and that the explanation of work in terms of thermodynamics has nothing to do with physics in the slightest?

I disagree. It may be confusing at first to relate seemingly unrelated topics together, but that's how MCAT passages are.

They force us to relate those topics together and filter through all our basic knowledge in order to find out what we really need to the questions.

Edit: Also, W = -PV = -(Pressure)*(Volume). Which could also fit into Gen Chem under Gases.

 

[tlaloc87]

I disagree. It may be confusing at first to relate seemingly unrelated topics together, but that's how MCAT passages are.

They force us to relate those topics together and filter through all our basic knowledge in order to find out what we really need to the questions.

Edit: Also, W = -PV = -(Pressure)*(Volume). Which could also fit into Gen Chem under Gases.

You mean you agree? I was saying that my explanation of work being KE from thermo can also help in understanding the definition and expression of work in physics. The OP clearly didn't see that, but, as you said, the MCAT will force you to relate topics. That is why it is helpful to be able to see the "big picture"--why would physics and chemistry use the same term ("work") for two completely unrelated concepts?

 

[g8orlife]

You mean you agree? I was saying that my explanation of work being KE from thermo can also help in understanding the definition and expression of work in physics. The OP clearly didn't see that, but, as you said, the MCAT will force you to relate topics. That is why it is helpful to be able to see the "big picture"--why would physics and chemistry use the same term ("work") for two completely unrelated concepts?

Gotcha! I misread your post.

 

[Phantastic]

(FYI: [d] = "delta" or "change in")

Einitial + Work = Efinal
thus (PEinitial + KEinitial) + W = (PEfinal+ KEfinal)

W = Fdcos(theta) = mad = [d]PE = mgh = 1/2k(x^2) = [d]KE = 1/2m(v^2) = Joules = [d]E = mc[d]T = fd = "mu" FNd = "mu" mgd = Newton*meter = (kg*m^2)/(s^2) = Power*time = (I^2)Rt = IVt = Watt*seconds ... etc.

 

[aladdin82]

W = P.E + K.E is correct
What I am saying is..

For instance, we can express Work done by spring

W = 0.5kx^2
as you know, x is displacement which is generated by some forces
and, k is spring constance.

Then we can find P.E if we've known the K.E. Let's assume K.E is Fd(d is distance between 2 applied points of forces, Of course, F is force)
Then we can write P.E is
P.E = 0.5kx^2-Fd

As we know, the shape of the equation is for evaluating P.E of Spring.
]
So, We can say tha W=P.E + K.E is correct.