Stuck on Physics Assignment: Find Force & Distance

[GFreeman64]

Hello everyone,

This is my first time in Physics forums so forgive me if I stray from the convention for asking questions. Me and a friend are completely stumped on our first assignment for the year and we are not sure where else to turn!

Homework Statement

A particle of mass m can move in a straight line. The only force acting on the particle opposes its motion, and depends only on its speed. The time T taken for the particle to come to rest after it is set into motion with speed u is given by the formula $T=klog(1+au)$ for all u, where k and a are positive constants. Find the force acting on the particle during the interval $0\leq t\leq T$ as a function of its speed v. Find also the distance the particle travels in time T.

Hint: Recall that the first fundamental theorem of calculus states that $\frac{\mathrm{d} }{\mathrm{d} x} \int_{a}^{x}f(w)dw=f(x)$.

Homework Equations

$T=klog(1+au)$, $\frac{\mathrm{d} }{\mathrm{d} x}, \int_{a}^{x}f(w)dw=f(x)$

$F(v)=m\frac{\mathrm{d}v }{\mathrm{d} t}$

The Attempt at a Solution

As far as I can see the best thing to do is take Newton's second law stated above, and separate the variables to give:

$m\int \frac{dv}{F(v)}=\int_{0}^{T}dt$

Which would give simply T on the RHS, so substituting for T would give:

$m\int \frac{dv}{F(v)}=klog(1+au)$

Now I am not sure how to proceed. I am tempted to try and take F(v) out of the integral to the other side of the equation, but I'm not sure whether this is allowed:

$m\int \frac{dv}{klog(1+au)}=F(v)$

If so, then my answer would simply be $\frac{mv}{klog(1+au)}=F(v)$. Any ideas as if this is correct?

For the second part I think I need to modify our starting point with Newton's second law to say:

$mv\frac{\mathrm{d} v}{\mathrm{d} x}=F(v)$

But after that I am a bit lost. If I have the equation for F(v) perhaps I can separate variables again and then express v as dx/dt and integrate again. Unfortunately i cannot see far enough ahead to see how this would work.

Thanks everyone in advance for taking a look at this!

 

[mathfeel]

The Attempt at a Solution

As far as I can see the best thing to do is take Newton's second law stated above, and separate the variables to give:

$m\int \frac{dv}{F(v)}=\int_{0}^{T}dt$

Which would give simply T on the RHS, so substituting for T would give:

$m\int \frac{dv}{F(v)}=klog(1+au)$

Now I am not sure how to proceed. I am tempted to try and take F(v) out of the integral to the other side of the equation, but I'm not sure whether this is allowed:

You are almost there. You cannot have a definite integral of the RHS, but an indefinite integral on the LHS. So the correct equation is:
$$m\int^{0}_{u} \frac{dv}{F(v)}=\int_{0}^{T}dt = klog(1+au)$$
Suppose you know $F(v)$, how would you do the integral? You would have to taking anti-derivative of $\frac{m}{F(v)}$. This should give you a hint of how it is related to the RHS. If still unsure, take derivative with respect to $u$ on both sides.

As for the second part, since $T(u)$ is the time it takes to stop the particle with velocity u, then at any intermediate time t when the particle is traveling with velocity $v(t)$, it would take $T(v)$ more time to stop it. Therefore, $t + T(v)= T$. This allow you to invert and find $v(t)$, from which you can integrate to find $x(t)$.

 

[GFreeman64]

Ah hah!

So if we were to differentiate both sides with respect to u, the RHS would become ka/(1+au). Then using the theorem given in the hint the LHS would become -1/F(u). So that gives us:

$F(u)=-\frac{1+au}{ka}$.

However the question asked for the force in terms of v, so how might I go about that from here?

Thanks very much for your help so far!

 

[*LouLou*]

Can T=K log(1+au) be rearranged to give us u?

u=1/a * ((10^(T/K)) -1))

If that is substituted back into the equation for force...

F(T)=-(10^T/K) / Ka

Does this get us anywhere?

 

[mathfeel]

However the question asked for the force in terms of v, so how might I go about that from here?

Let me give you a simpler example here. Suppose $f(x)$ satisfies $\int_0^y f(x) dx = \frac{y^2}{2}$. I differentiate both side with respect to $y$ and get $f(y) = y$. What is $f(x)$?

 

[GFreeman64]

Hmm, so in your example here, if f(y)=y then it would also be true that f(x)=x. Is then as simple as saying that if:

$F(u)=-\frac{1+au}{ka}$ then $F(v)=-\frac{1+av}{ka}$?

If we substitute this expression for F(v) into the original integral it seems to almost work, apart from the fact that we still have a factor of m:

$m\int_{u}^{0}\frac{dv}{F(v)}=kmlog(1+au)$

Thanks again.

 

[GFreeman64]

If we were to say that:

$F(v)=\frac{m(1+av)}{ka}$

Then the integration seems to work and is equal to $klog(1+au)$. For now I'll assume this is correct.

For the second part I have taken the equation t+T(v)=T(u) to give:

$t+klog(1+av)=klog(1+au)$

Rearranging this for v gives:

$v=\frac{1+au}{a10^{t/k}}-\frac{1}{a}$

Integrating this between 0 and T gives:

$x=-\frac{k(1+au)10^{-t/k}}{a}-\frac{t}{a}$

Which is showing a negative value for the position x. I think this is probably incorrect, but I am not sure where I have gone wrong...