With what force does the cobra press on the ground when it is preparing to jump, rises vertically upwards at a constant speed v? The mass of the snake is m, its length is l.

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    <h3 id="back-link"><a href="/#2.2">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.2.42.$ With what force does the cobra press on the ground when it is preparing to jump, rises vertically upwards at a constant speed $v$? The mass of the snake is $m$, its length is $l$.
    </p>

    <h3>Solution</h3>
    <p>
        

Let's consider small part of cobra with the length $\Delta x$ and mass $\Delta m$:
\begin{equation*} \Delta P = \Delta m \cdot v \end{equation*}
Since the snake is homogeneous
\begin{equation*} \Delta m = \frac{m}{L} \cdot \Delta x \end{equation*}

By newton's second law:
\begin{equation*} F=\frac{\Delta P}{\Delta t} = \frac{\Delta m}{\Delta t} \cdot v \end{equation*}

$$F = \frac{m}{L} \cdot \frac{\Delta x}{\Delta t} \cdot v = \frac{mv^2}{L}$$

By definition of speed
\begin{equation*} \frac{\Delta x}{\Delta t} = v \end{equation*}

Whence the force associated with the acceleration of the centre of mass
\begin{equation*} F = \frac{\Delta P}{\Delta t} = \frac{mv^2}{L} \end{equation*}

Let's take everything together and find the support reaction force
\begin{equation*} N = mg + \frac{mv^2}{L} = m\left(g + \frac{v^2}{L}\right) \end{equation*}

\begin{equation*} \boxed{N = m\left(g + \frac{v^2}{L}\right)} \end{equation*}

    </p>

    <h4>Answer</h4>
    <p>
        $$N = m\left(g + \frac{v^2}{L}\right)$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">   
      Almaskhan Arsen<br>
    </p>


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