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I'm given the following problem: enter image description here

Now, the following is my attempt at a solution: enter image description here

I have two problems: (1) With my equation for $v$, I end up having to take the log of $0$, which is obviously undefined, so something's gone wrong here.

(2) Secondly, I 'establish' that $ W \neq \Delta T $ (i.e. the work done is not the change in kinetic energy, contrary to the Work-Energy principle.

(Also, the reason I've printscreened my answer is that it would take a very long time for my to type in $\LaTeX$).

Could someone help me see the light? Thanks in advance.

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  • $\begingroup$ Why did you substitute $dv/dt = v\, dv/dx$? Just integrate both sides of $-\mu R = m {dv\over dt}$ with respect to time. $\endgroup$ Mar 25, 2014 at 22:02
  • $\begingroup$ Ah. That seems quicker. But my way should work, right, since $\frac{dv}{dt}=\frac{dv}{dx} \times \frac{dx}{dt}$ and $v:=\frac{dx}{dt}$. $\endgroup$
    – beep-boop
    Mar 25, 2014 at 22:03
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    $\begingroup$ Don't you need $\eta\, d\eta$ in your first integration, not just $d\eta$? $\endgroup$ Mar 25, 2014 at 22:07

2 Answers 2

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It was interesting to follow your approach through to the end.

As a starting point, fix the first integral so it says $$\int_0^x{\left(-{\mu R\over m}\right)\,d\xi} = \int_{v_0}^v{\eta\,d\eta}\,.$$ Normally we'd save on the algebra by abbreviating ${\mu R\over m} \equiv a$. But since you're doing an unusual approach, I'll carry around the symbols so we can see how energy and kinematics relate.

Performing the integration, we have $$-\left({\mu R\over m}\right)x={1\over 2}(v^2-v_0^2)$$ or $$v = \sqrt{v_0^2-{2\mu R\over m}x}\,.$$ From this we find the stopping distance by setting $v = 0$ and solving for $x$; call the result $d$: $$d = {mv_0^2\over 2\mu R}\,.$$

This also shows that the work done by friction is equal to the change in kinetic energy; for the work done by friction is $W = \int{{\bf F}\cdot d{\bf s}} = -\mu R\int{ds} = -\mu R d$, and using the result for $d$ above we have $W = -{1\over 2} mv_0^2$.

Next let's find the trajectory $x(t)$. We have $dt = {dx\over v}$ so that $$\int_0^x{d\xi\over\sqrt{v_0^2-{2\mu R\over m}\xi}} = \int_0^t{dt^\prime}\,.$$

Substitute $u = v_0^2 - {2\mu R\over m}\xi$, turn the crank, and the equation becomes $${m\over 2\mu R}\int_{v_0^2-{2\mu R\over m}x}^{v_0^2}{du\over\sqrt{u}} = t\,.$$

The integration is trivial, so we find $$v_0 - \sqrt{v_0^2-{2\mu R\over m}x} = {\mu R\over m}t$$ which can be solved for $x$ to yield $$x = v_0 t - {1\over 2}\left({\mu R\over m}\right)t^2\,.$$ The horizontal velocity is $dx/dt$, or $$v_x = v_0 - \left({\mu R\over m}\right) t\,.$$ Solve for the time $t_\star$ when $v_x = 0$: $$t_\star = {mv_0\over \mu R}\,.$$

Note that $t_\star$ is the initial momentum divided by the magnitude of the stopping force.

This is by no means the way I would have solved the problem myself, but as I say it was interesting! I think that a person who is just learning energy methods would benefit from doing it this way so that each step could be compared to the familiar kinematical approach.

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I'm not allowed to comment, but as a follow up to Jason Zimba: $$ \frac{d}{dx}(v^2) = 2v\frac{dv}{dx} $$ So a factor of 1/2 is missing if we follow the route $$ \int_{x_0}^x v(\eta)\,\frac{dv}{d\eta}\,d\eta = \frac{1}{2}\int_{x_0}^x\frac{d}{d\eta}(v^2)\,d\eta = \frac{1}{2}[v(x)^2-v(x_0)^2] = \frac{1}{2}(v^2-v_0^2) \,. $$

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