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

[Ocata]

Does this derivation:

...imply:

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

 

[Ocata]

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.

 

[256bits]

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).

My best guess is that x(t) ≠ t
So I would also guess that F(x(t)) ≠ F(t)

Hmmmm

 

[Ocata]

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:

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

 

[Svein]

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)).

 

[Ocata]

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≠ t^{4} , 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) = t^{4}. 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 = t^{4}

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

 

[Nathanael]

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.

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).

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)

 

[Nugatory]

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.

 

[DrGreg]

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$$
F_2(t) = F_1(x(t))
$$

 

[Ocata]

Thank you,

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

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) = t^{4} and arriving at a force with respect to time: f(t) = 60t^{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)) ?

My first guess is something like this:

 

[Svein]

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 x(t)=Bt^{4}. For x≥0, this is one-to-one, so t=(\frac{x}{B})^{-4}.

 

[Ocata]

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...