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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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


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.


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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