Let's say I have an arbitrary path from point $A$ to point $B$. A particle moves from $A$ to $B$ under the influence of gravity but also impeded by a frictional force $\vec{F_\mu}$. The gravitational field does work on the particle such that $$ \int_a^c{\vec{F_g} \cdot \mathrm{d}\vec{x}} ~=~ W ~=~ \Delta T \tag{1} \,,$$ and on the other side because the frictional force is always parallel to the path, $$ \int_a^c {\vec{F_\mu}} \, \mathrm{d}\vec{x} ~=~ -W \,.$$ So should it not be true that $$ \int_a^x \left(\vec{F_g} \cdot \hat{T}-\vec{F_\mu}\right) \, \mathrm{d}x ~=~ \Delta T \tag{2} $$ is the change in kinetic energy between points $a$ and $x$ on the path? In that case, by incorporation of arclength and rearranging the kinetic energy to be in terms of velocity is it not true that $$ \Delta t ~=~ \int\limits_a^b{\sqrt{\frac{1+{y'}^{2}}{\vec{F_g} \cdot \hat{T}-\vec{F_\mu}}}} \, \mathrm{d}x \tag{3} \,,$$(integrating the $t=d/s$ equation essentially)? If this is the case, then for the path $y=-x+10$, ${y'}^2 = 1$, so $$ \Delta t ~=~ \int\limits_0^{10}\sqrt{\frac{2}{\frac{F_g\sqrt{2}}{2}}} \, \mathrm{d}x ~=~ \int\limits_0^{10}{\sqrt{\frac{4}{\sqrt{2}F_g}} \, \mathrm{d}x} ~=~ 5.3723 \, \mathrm{s} \tag{4} \,.$$

This blatantly conflicts with my HS level physics answer of $2.02 \, \mathrm{s} ,$ which (while hastily obtained) seems more plausibly the correct answer.

Question: How can my first method be adapted to obtain the correct time for the movement to occur, and is it a proper method for more convoluted paths? Also please point out the probably obvious error which is causing this issue.


The solution in your case is:

For conservative system (no friction ) the energy $E$ is conserved.


Thus: $$E_i=E_f \mapsto T_i+U_i=T_f+U_f\tag 1$$

where $T$ is the kinetic energy and $U$ the potential energy. i; initial state and f; final state


$T=\frac{1}{2}\,m\,\vec{v}\cdot \vec{v}$



$\vec{v}=\frac{dx}{dt}\left[ \begin {array}{c} {1}\\ \left( {\frac {d}{dx}}y \left( x \right) \right) {}\end {array} \right] $


$$E(x,xp)=T+U=1/2\,{{\it xp}}^{2}m+1/2\, \left( {\it ys} \left( x \right) \right) ^ {2}{{\it xp}}^{2}m+mgy \left( x \right) \tag 2$$

where : $xp=\frac{dx}{dt}$ and $ys=\frac{dy}{dx}$

for the

initial state: $x=0\,,xp=0$ and the final state: $x=L\,,xp=dx/dt$ we get with equations (1)

$$E_i(0,0)=E_f\left(L,\frac{dx}{dt}\right)\tag 3$$

with equation (2) we solve equation (3) for $dx/dt$ and get:

$$\frac{dx}{dt}={\frac {\sqrt {2}\sqrt { \left( 1+ \left( {\it ys} \left( L \right) \right) ^{2} \right) g \left( y \left( 0 \right) -y \left( L \right) \right) }}{1+ \left( {\it ys} \left( L \right) \right) ^{2}}} \tag 4$$

we solve equation (4) for $t(x=L)$ and get:

$$\boxed{t(x=L)=\frac{1}{2}\,{\frac {L \left( 1+ \left( {\it ys} \left( L \right) \right) ^{2 } \right) \sqrt {2}}{\sqrt { \left( 1+ \left( {\it ys} \left( L \right) \right) ^{2} \right) g \left( y \left( 0 \right) -y \left( L \right) \right) }}}} $$

your example:


$ys(x)=\frac{d y(x)}{dx}=-1$


$$t=\frac{L}{\sqrt{g\,L}}=\frac{10}{\sqrt{9.81\,,10}}\approx 1 [s]$$


I) If you have a friction force, to get the time $t$ you must solve the equation of motion (numerically) and stop the simulation when x reach the length $L$.

II) The general description of a path is $x=x(s)$ and $y=y(s)$ where s is the path parameter (path length), but the steps to obtain the time $t$ are the same.


I am fairly certain that you can't write the force as a derivative of a gradient when there exists a non-conservative force(friction here).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.