Ascending a Sinusoidal Wire

Due to the fact that the gravitational field is a conservative field, the $\Delta E$ of energy does not depend on a particular walk of the point. So, we can write, for the 'Law of conservation of Energy', that

$\displaystyle \Delta E_{f,i}=0$

Where $E_i$ is the energy of the point at the origin, while $E_f$ is the energy when it stops. Let's fix at $z=0$ the potential energy $E_p=0$ . So at the beginning, the material point has just cinetic energy

$\displaystyle E_i=\frac{1}{2}m v_{i}^2$

At the end, when $v_f=0$ , it has just the amount of potential energy

$\displaystyle E_f=mgz_f$

So

$\displaystyle mgz_f-\frac{1}{2}mv_{i}^2=0 \text{,} \quad z_f=9.8 m$

We have to find the parametrization of $\vec{P}$ . Vectors $\vec{u_1}=(1,2,3)$ and $\vec{u_2}=(-1,5,-3)$ give us directions, but we need versors. So let's multiply each vector by the opposite of its module.

$\displaystyle ||\vec{u_1}||^{-1}=\frac{1}{\sqrt{14}} \text{,} \quad ||\vec{u_2}||^{-1}=\frac{1}{\sqrt{35}}$

$\displaystyle \vec{u_1}=(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}) \text{,} \quad \vec{u_2}=(-\frac{1}{\sqrt{14}},\frac{5}{\sqrt{14}},-\frac{3}{\sqrt{14}})$

And so $\vec{P}$ becomes

$\displaystyle \vec{P}(\alpha)=\left(\frac{\alpha}{\sqrt{14}}-\frac{2\sin{\alpha}}{\sqrt{35}},\frac{2\alpha}{\sqrt{14}}+\frac{10\sin{\alpha}}{\sqrt{35}},\frac{3\alpha}{\sqrt{14}}-\frac{6\sin{\alpha}}{\sqrt{35}}\right)$

Since we previously calculated $z_f$ , we know the $z$ -coordinate of the ending point. So

$\displaystyle 9.8=\frac{3\alpha}{\sqrt{14}}-\frac{6\sin{\alpha}}{\sqrt{35}} \text{,} \quad \alpha=10.958$

Eventually

$\displaystyle \vec{P_f}=\vec{P}(\alpha=10.958)=(3.266,4.168,9.8)$

The distance of $\vec{P_f}$ from the origin is

$\displaystyle ||\vec{P_f} ||=11.139m$