# Newtonian Mechanics – Inconsistent Result from the Kinetic Energy Theorem Explained

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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?

(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.