A man in standing on the edge of a flat surface, $H^2$ meters above the ground ($H>0$), which is also the highest and starting point of a parabolic skiing course, given by the simple parametrization:
$$\gamma(t)=(t,t^2)\\t\in[-H,h]$$
When $h>0$. At some moment he begins to ski, reaching the height $h^2<H^2$ twice: once before reaching the minimum point of the course (which is at ground level), and once again afterwards. We will define the second point he reaches to be the endpoint of the course.
It is given that the course is not smooth, and the man feels a friction force $\vec{F}$ along the course. At every point of the course, the force is parallel to the parabola, and its orientation is opposite to the direction of the man. The force at every point of the course is given by $|\vec{F}|=\mu|\vec{N}|$, when $\mu\in\mathbb{R}$ is a given constant, and $\vec{N}$ is the normal force at that point.
I am required to compute the work of the friction force along the course.
I don't have a problem with the physics of the problem (I hope), but with the math. I'll explain:
I know that at every point $(t,t^2)$ of the course, the normal force $\vec{N}$ is given by $|\vec{N}|=mg\cos\alpha$, when $\alpha$ is the angle between the $\hat{x}$ axis and the tangent line to the parabola. Therefore, at every point of the course, $\tan\alpha$ is given by $\tan\alpha = \frac{d}{dt}t^2=2t$. The direction of $\vec{F}$ would be, therefore, $-\cos\alpha\hat{x}-\sin\alpha\hat{y}$. In conclusion, the friction force is given by:
$$\vec{F}=-\mu mg\cos\alpha\left(\cos\alpha,\sin\alpha\right)=-\mu mg\left(\frac{1}{1+4t^2},\frac{2t}{1+4t^2}\right)$$
Now:
$$W_F\equiv\int_\gamma\vec{F}\ d \vec{r}=-\mu mg\int_{-H}^{h}\left(\frac{1}{1+4t^2},\frac{2t}{1+4t^2}\right)\cdot(1,2t)\ dt=-\mu mg\int_{-H}^{h}dt=-\mu mg(h+H)$$
Something seems odd here - how come $W_F$ is given by the the product of the size of the force, with the horizontal length of the curve? This would seem legitimate to me if the course was horizontal, but it's not. What am I missing here? Is my math incorrect, or maybe is it my physics?
Thank you!
The calculation appears to be correct. The result seems more intuitive when you realise that $h+H$ is also the time taken to traverse the course. Since the infinitesimal work at any given time is constant, the total work should be proportional to the time taken. Interestingly, you get the same result for any path of the form $\gamma(t)=(t,f(t))$ for some differentiable function $f$, which is due to the force having the property that $\vec{F} \propto \frac{\vec{v}}{||v||^2}$.