Inconsistent result from the kinetic energy theorem I have a simple physics 101 problem that leads to a strange result.
A cart of mass $m$ is bound to move without friction on a rail placed in the vertical plane (as shown in the picture).
The rail is formed by two circular pieces of radius $R$ glued together.

Initially, the cart is standing still at the bottom of the rail, and at time $t=0$, constant force $\mathbf{F}$ is applied to the cart. The force $\mathbf{F}$ is at each point of the trajectory tangent to the rail.
Once the cart reaches a height of $2R$, the force $\mathbf{F}$ disappears.
Question: what is the minimum value of $F$ that makes the cart reach a height equal to $2R$?
Solution
An approach to solve the problem would be to use the work-energy theorem, which states that the variation of kinetic energy is equal to the work done
$$W_{o\to b} = \frac{1}{2}m v^2(b) - \frac{1}{2}m v^2(o) $$
The work done by the two forces (the weight and the force $\mathbf{F}$) can be computed as
$$W_{o\to b} = \int_o^b (\mathbf{F}+m\mathbf{g})\cdot \mathrm{d}\mathbf{s}=F\int_o^b\mathrm{d}s - mg2R=F\pi R - 2mgR,$$
where we used the fact that the force $\mathbf{F}$ is always parallel to the displacement $\mathrm{d}\mathbf{s}$.
On the other hand, I know that the velocity in $o$ is zero, and if I want to compute the minimum value of the force $\mathbf{F}$ that allows the cart to reach the height $2R$, I assume that the velocity is zero also in the point $b$, hence
$$\frac{1}{2}m v^2(b) - \frac{1}{2}m v^2(o)= 0-0 =0$$
Therefore the work-energy theorem becomes
$$W_{o\to b} = F\pi R - 2mgR = 0,$$
which, solving for $F$, gives
$$F= \frac{2}{\pi} mg.$$
This result seems puzzling to me.
Indeed, the beginning of the rail is vertical, so the force $F$ needed to overcome the weight and allow the cart to move up must be at least $mg$.
Why does the work-energy theorem lead to the wrong result? Did I apply it in the wrong way?
 A: Since this is a 1-D problem, we can define a potential energy due to the applied force as well, given by
$$
U_F \equiv -\int_0^\vec{r} \vec{F} \cdot d\vec{r} = - F s
$$
where $s$ is the arc length along the path.
Adjusting the value of $F$ can just be thought of as adjusting the "strength" of this potential energy.
A bit of geometry shows that we can relate $s$ to the height $y$ by
$$
s = R \begin{cases} \arcsin(y/R) & 0 \leq y \leq R \\ \arccos(2 - y/R) + \pi / 2 & R < y \leq 2R\end{cases} 
$$
and so the total potential energy for the object is
$$
U = m g y - F R \begin{cases} \arcsin(y/R) & 0 \leq y \leq R \\ \arccos(2 - y/R) + \pi / 2 & R < y \leq 2R \end{cases}.
$$
For $F = mg$, the potential energy looks like this:

It is easy to see that a particle at $y = 0$ with negligible initial kinetic energy will make it to $y = 2R$ (and will have a substantial amount of KE when it gets there.)  On the other hand, if we plot the potential for $F = 2 m g /\pi$, we get the following graph:

If the cart did get to $y = 2R$ in this potential, it would arrive there with the same KE it started with;  we would have $\Delta K = -\Delta U = 0$ between these points.  But it would have to have an initial velocity that's sufficient to get it "over the hump".  On the other hand, if the cart starts with zero KE, it can't get over the hump and so will never arrive at $y = 2R$.
A: The equation of motion is:
$$m\,\ddot s+F-\cos\left(\frac sR\right)\,m\,g=0$$
form here with $~s=\pi\,R~$ and $~\ddot s=0~$ you obtain
that
$$F=-m\,g$$
now your problem
multiply the EOM with $~\dot s~$  and
integrate you obtain
$$\frac m2\dot s^2=-\int F\,ds-\int \cos\left(\frac sR\right)\,m\,g\,ds=-F\,s-m\,g\,R\sin\left(\frac sR\right)\quad\Rightarrow\\
W=-F\,s-m\,g\,R\sin\left(\frac sR\right)$$
but with $~s=\pi\,R~$ you obtain  that $~F=0$ thus your ansatz is probably  wrong

but if the work is
$$W=m\,g\,y+F\,y$$
you obtain for y=$~2\,R~,W=0$ ,$~F=-m\,g~$
A: The way that the problem is stated, $F >= mg$ for the cart starts to move. But afterwards, the force will always be bigger than the tangential component of mg. So the cart will accelerate all the time. So the minimum is $F = mg$
And if not only the magnitude but also the direction of $\mathbf F$ is constant, but the cart is constrained to follow the trajectory?
$\mathbf F_{net} = m\mathbf a$
During the trajectory, there are normal forces. As they are always orthogonal to the allowable displacement, we can get rid of them by making a dot product with the elementary displacement:
$(\mathbf F - m\mathbf g)\mathbf{.dr}  = m\mathbf a\mathbf{.dr}$
Until $x = R$, naming $cos(\theta) = 1 - \frac{x}{R}$:
$$(F - mg)drcos(\theta) = m\frac{\mathbf {dv}}{dt}\mathbf{.dr}\implies (F - mg)Rd\theta cos(\theta) = m\mathbf v\mathbf {.dv} = d\left(\frac{1}{2}mv^2\right)$$
Integrating from $0$ to $\frac{\pi}{2}$, and supposing $v = 0$ for $\theta = 0$:
$$(F - mg)R = \frac{1}{2}mv_1^2 $$ It is necessary that $F >= mg$ for this equation be fulfilled
The second path is similar, except that $cos(\theta) = \frac{x}{R} - 1$. When $\theta = -\frac{\pi}{2}$,  $v = v_1$ and the integral is from $-\frac{\pi}{2}$ to $0$
$$(F - mg)R = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2 $$
If $F = mg$, $v_1 = 0$, and $v_2 = 0$.
So, also in this case, $F = mg$
But the problem makes more sense. The speed in this limit situation is always zero. The projection of F and mg will always be equal, so given a small initial velocity, it keeps the same.
A: (a) If the force were constant neither its direction nor its magnitude would change. We are assuming here that the magnitude is constant.
(b) It isn't the work-energy theorem that has generated a paradox, but your assumption that for minimum $F$ the cart's KE is zero at b. The energy condition that we do have a right to impose is the less stringent one that $\text{KE} ≥ 0$ at b. Therefore, from your work calculations, $F ≥ \frac 2\pi mg$.
(c) As you've pointed out, the equality, $F = \frac 2\pi mg$ cannot be the right choice, because it wouldn't allow the cart to get started on the climb! We know how large $F$ does need to be.
(d) [added at the suggestion of YiFan] With $F = mg$ (the minimum value of $F$ for the journey to start) the cart will gain KE throughout the journey (except at the very beginning and the very end), so its KE at b can't be zero. The general condition for gaining KE at any point along the curve is easily shown to be $\sin \theta ≤\frac F {mg}$ in which $\theta$ is the local angle of the curve to the horizontal.
(e) Would the problem have a less trivial answer if the left hand and right hand pieces of the curve were exchanged, so that the vertical bit came in the middle of the complete curve? [I believe that it has!]
