Work-Energy Derivation question: F(x(t)) = F(t) ?

In summary: F_1(x(t))$$and no confusion arises. This is a lot of pedantry, but worth keeping in mind.In summary, the conversation is discussing a derivation that involves functions of time and position, and how they relate to each other. The main points are that x(t) is a function of time and F(x) is a function of position, so F(x(t)) means that F is
  • #1
Ocata
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Does this derivation:

work-energy_zpsflypwpst.png


...imply:

work-energy2_zpszo01eiy4.png
My best guess is that x(t) ≠ t
So I would also guess that F(x(t)) ≠ F(t)

But then how can this derivation be explained?

How can F(x(t)) = m(a(t))? What does that actually mean?
How come it's not: F(x(t)) = m(a(x(t))) ? Why/How does the x just cancel out? Thank you
 
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  • #2
Would this question be better suited for the math section? If so, please feel free to transfer it. It seems to be bordering on a math related question within a physics derivation and I don't think I'll fully understand the derivation without understanding how the math is being work at this part.

Again, thanks.
 
  • #3
Ocata said:
How can F(x(t)) = m(a(t))? What does that actually mean?

x(t) implies x is a function of t, ie for a particular t there is a particular x.
F(x) implies that F is a function of x
F(x(t)) implies ... You should be able to figure that out.
Same for a(t).

Ocata said:
My best guess is that x(t) ≠ t
So I would also guess that F(x(t)) ≠ F(t)
Hmmmm
 
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  • #4
256bits said:
x(t) implies x is a function of t, ie for a particular t there is a particular x.
F(x) implies that F is a function of x
F(x(t)) implies ... You should be able to figure that out.
Same for a(t).Hmmmm

The following might describe more precisely where I'm experiencing some uncertainty,

Suppose I start with some object of 5kg moving in some way described by x(t) = t^4, then:

7f70758a-b644-4a88-b672-4c6fce7a6af0_zpslqatzu2p.jpg


Why do I arrive at 60t^2 for f(t)
and arrive at 60t^6 for f(x(t))?

It seems like f(t) = 60(t)^2
while f(t(x)) = f(t^4) = 60(t^4)^2 = 60t^6
 
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  • #5
You need to fix up your formula. If x is a distance and t is time you cannot have x=t4 (the dimensions do not match). You could get away with x = Ct4, but then the dimension of C would be m/s4.

As to the rest of your questions: Of course you cannot have F(t) = F(x(t)). The dimensions do not match! If you have a well-behaved function, you can possibly invert the x(t) to t = g(x), and then you will have F(t) = F(g(x)).
 
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  • #6
Svein said:
You need to fix up your formula. If x is a distance and t is time you cannot have x=t4 (the dimensions do not match).

Yes, x is a distance and t is time, so x≠t and x≠ [itex]t^{4}[/itex] , however, I am convinced I included parenthesis everywhere to indicate that x is a function of t. That is, x(t) = t and x(t) = [itex]t^{4}[/itex]. As a function, the dimensions don't have to match, right? That is, displacement is a function of time, then x(t) = t can be written. I do not see where I may have written x = t or x = [itex]t^{4}[/itex]
Svein said:
As to the rest of your questions: Of course you cannot have F(t) = F(x(t)). The dimensions do not match! If you have a well-behaved function, you can possibly invert the x(t) to t = g(x), and then you will have F(t) = F(g(x)).

Okay, and if I cannot have F(t) = F(x(t)), then why does the formula in the original post say: F(x(t)) = m(a(t)) = m v'(t) ? <---this is what is written in my book.

So I'm ending up at a contradiction. On one hand, F(x(t)) ≠ F(t), but then my book states F(x(t)) = m(a(t)). Can this be?

Thank you
 
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  • #7
Ocata said:
As a function, the dimensions don't have to match, right?[/itex]
When you're speaking of an equation which is physically meaningful, the dimensions must match. With that said, there is no need to worry about dimensions when contriving a particular example like this, because you can assume a particular unit system which leaves the dimensional-constant with unit value.
Ocata said:
On one hand, F(x(t)) ≠ F(t),
You could say F(x(t))=F'(t) where F' is some other function (not the same as F unless x=t).
Ocata said:
but then my book states F(x(t)) = m(a(t)). Can this be?
That's what I see it as saying: some function F of x of t is equal to (m times) some other function a(t).

It's really not an interesting statement (and isn't relevant to the Work-KE theorem).

It also doesn't imply a(x(t))=a(t) (which is the source of your contradiction).P.S. (t4)2=t8 (not t6)
 
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  • #8
In your example, you went wrong in the step where you wrote $$F(x(t)) = m(a(x(t))) = 60[x(t)^2]^2$$ The first equality is correct, the second is not. The problem is that ##a(t)## and ##a(x)## are different functions: the first is the formula you use to calculate the acceleration at a given time and the second is the formula you use to calculate the acceleration when the object is at a given position and is not equal to ##12t^2##. Whenever you see ##a(t)## you are allowed to substitute ##12t^2##, but you cannot make the same substitution when you see ##a(x)##.

The same is true of ##x##, ##v##, and ##F##; for example ##x(t)=t^4## but ##x(x)=x##.

If you work through your example again but use different names (for example ##F_x(x)## and ##F_t(t)## for the force) for the functions of time and of position it will all come out consistently.
 
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  • #9
Many physics authors and teachers use a convention where, for example, ##F(x)## denotes a force as a function of distance, and ##F(t)## denotes the same force as a function of time. Strictly speaking, from a pure mathematician's point of view, this is nonsense and you should really use different names for the two functions, e.g. ##F_1(x)## to denote force as a function of distance, and ##F_2(t)## to denote force as a function of time. With this notation we have$$
F_2(t) = F_1(x(t))
$$
 
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  • #10
Thank you,

If I were to look at each function graphically, I have something like this:

f8114d58-f86d-4867-a951-7e260ee09e39_zpsvyhofmjx.jpg


However, I'm having some difficulty in figuring out how to express the force as a function of distance.

From starting with a displacement as a function of time: x(t) = [itex] t^{4} [/itex] and arriving at a force with respect to time: f(t) = [itex]60t^{2}[/itex]

How can I take it one step further and find a function of force with respect to displacement? F(x) or F(x(t)) ?

My first guess is something like this:

1d9bd928-e053-40e4-a07e-f3d5b0c46cae_zpsiq8gqcmg.jpg
 
  • #11
Ocata said:
From starting with a displacement as a function of time: x(t) = t4 t^{4} and arriving at a force with respect to time: f(t) = 60t260t^{2}

How can I take it one step further and find a function of force with respect to displacement? F(x) or F(x(t)) ?
Well, first we introduce a dimension-correcting factor B (with dimension m/s4), so that [itex] x(t)=Bt^{4}[/itex]. For x≥0, this is one-to-one, so [itex]t=(\frac{x}{B})^{-4} [/itex].
 
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  • #12
Thank you Svein. I've continued to look into this and have made good progress. I must go on a tangent and ask a new question regarding momentum before coming back to the topic of kinetic energy. To be continued...
 

1. What is the meaning of F(x(t)) = F(t)?

This equation represents the force applied to an object at a specific time (t) and position (x). It shows that the force is independent of the object's position and only depends on the time at which it is applied.

2. How is F(x(t)) = F(t) derived?

This equation can be derived from the work-energy theorem, which states that the work done on an object is equal to its change in kinetic energy. By using the definition of work and substituting in the equation for force, we can arrive at F(x(t)) = F(t).

3. Can F(x(t)) = F(t) be applied to all types of forces?

Yes, this equation is applicable to all types of forces as long as they are independent of the object's position.

4. How does F(x(t)) = F(t) relate to the conservation of energy?

This equation is a consequence of the conservation of energy principle, which states that energy cannot be created or destroyed, only transferred from one form to another. F(x(t)) = F(t) shows that the force applied to an object does not depend on its position, indicating that the work done on the object is independent of its path and therefore, the total energy remains constant.

5. Can F(x(t)) = F(t) be used to calculate the work done on an object?

Yes, this equation can be used to calculate the work done on an object by integrating the force over time, as shown in the work-energy theorem.

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