I got a nonlinear equation, which describes the magnitude of a force as a function of time, but I don't know how to calculate the work done by the force. Given: $$F(t) = kv(t)$$ with $$v(t)= \left(1-\frac{k}{m}\right)^t v_{0}$$ and I want to derive: $$W = \int F(t) dx$$ and also want to convert $F(t)$ to $F(x)$
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2$\begingroup$ Welcome to this community! Recall that the work can be written (and is even more appropriately defined) as $$\int_{t_0}^{t_1} \pmb{F}(t) \cdot \pmb{v}(t) \ \mathrm{d}t\ .$$ $\endgroup$– pglpmCommented Dec 16, 2020 at 14:44
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2$\begingroup$ same $dx\rightarrow \dfrac{dx}{dt}dt$ $\endgroup$– EliCommented Dec 16, 2020 at 14:46
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$\begingroup$ Oh, yes. Indeed! Thanks. So, how about F(x). Can I transform F(t) to F(x)? $\endgroup$– Geika KiyomizuderaCommented Dec 16, 2020 at 14:48
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$\begingroup$ No, $F$ won't be a function of $x$, but a derivative of it, since the $F$ given depends on $v$. However you don't need that to compute the work. If you are actually interested in $x(t)$ you will have to solve a diff.eq $\endgroup$– ohneValCommented Dec 16, 2020 at 14:55
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1$\begingroup$ @ohneVal I disagree with your statement for F not being a function of x. F will become a function of x if you can write v as a function of x $\endgroup$– SteelCubesCommented Dec 16, 2020 at 14:58
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1 Answer
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$$ \begin{align}F(t)&=\space kv(t) \\
v&= \frac{dx}{dt}=(1-\large{\tfrac{k}{m}})^{\small{t}}\, v_0\\
{\rm d}x&=(1-{\tfrac{k}{m}})^{\small{t}} v_0 {\rm d}t\\
\int_0^x {\rm d}x&= \int_0^t (1-{\tfrac{k}{m}})^{\small{t}}\, v_0 {\rm d}t\\
x&=\frac{(1-{\frac{k}{m}})^{\small{t}}-1}{\ln(1-{\frac{k}{m}})} v_0\\
(1-{\tfrac{k}{m}})^{\small{t}} v_0&= v(t)= v_0 + x \space \ln(1-{\tfrac{k}{m}})\\ \end{align}$$
Substituting the values in the expression for $F(t)$ $$ \begin{aligned}
F(x)=k \left[v_0 + x \space \ln(1-{\tfrac{k}{m}})\right]
\end{aligned}$$
I hope you can calculate forward.
Regards.
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1$\begingroup$ A big thank you! I appreciate you! The last calculation is beautiful... $\endgroup$ Commented Dec 16, 2020 at 15:04
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$\begingroup$ @SteelCubes $~x={\frac { \left( 1-{\it kk} \right) ^{t}{\it v0}}{\ln \left( 1-{\it kk } \right) }} ~$ where $kk=\frac km~$ from hear you can obtain $~t=t(x)$ and substitute this in $~F=k\,v(t)\mapsto \,k\,v(t(x)) $ you obtain $F(x)$ $\endgroup$– EliCommented Dec 16, 2020 at 17:06
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2$\begingroup$ @Eli Firstly, it is $x= \frac{(1-\frac{k}{m})^t -1}{ln(1-\frac{k}{m})}v_0$. I think you forgot to put the lower limit of t=0 where x=0. And secondly, that's exactly what I did. I mean, I just took a few step forwards to reduce the calculation part. But I did exactly what you're saying. $\endgroup$ Commented Dec 16, 2020 at 18:41