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

$$W_{o\to b} = \int_o^b (\mathbf{F}+m\mathbf{g})\cdot \mathrm{d}\mathbf{s}=F\int_o^b\mathrm{d}\mathbf{s} – mg2R=F\pi R – 2mgR,$$

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?

## Best Answer

(a) If the force were constant neither its direction nor its magnitude would change. We are assuming here that the

magnitudeis 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 $F$

doesneed 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!]