Low-Speed Lorentzian Physics

We are given that mass of an object varies with its velocity as: $m = \frac{m_0}{\sqrt{1-v^2/c^2}}$ Also we know that force is rate of change of momentum i.e. $F = \frac{d}{dt}mv$ So we write force as: $\begin{aligned} F &= \frac{d}{dt} \frac{m_0 v}{\sqrt{1-v^2/c^2}} \\ &= \frac{m_0}{(1-v^2/c^2)^{3/2}} \frac{dv}{dt} \end{aligned}$ replacing $\frac{dv}{dt}=a$ ,We get $\begin{aligned} a &=\frac{\left(1-v^2/c^2\right)^{3/2}}{m_0} \times F \\ \end{aligned}$ Substituting, $\begin{aligned} v= \SI{5}{\meter\per\second} \\ F=\SI{1}{\newton} \\ m=\SI{1}{\kilo\gram} \\ c= \SI{10}{\meter\per\second} \end{aligned}$
We get $\begin{aligned} \large{a} &=\boxed{\SI[per-mode=symbol]{0.65}{\meter\per\second\squared}} \\ \end{aligned}$

Let $c = 10 m/s$

$F = m_{0} \dfrac{d}{dv} (\dfrac{v}{\sqrt{1 - \dfrac{v^2}{c^2}}}) \dfrac{dv}{dt} =$ $$

$m_{0} * (\dfrac{v^2}{c^2 * (1 - \dfrac{v^2}{c^2})^{\dfrac{3}{2}}} + \dfrac{1}{(1 - \dfrac{v^2}{c^2})^{\dfrac{1}{2}}}) \dfrac{dv}{dt} =$ $$

$\dfrac{m_{0}}{(1 - \dfrac{v^2}{c^2})^{\dfrac{3}{2}}} \dfrac{dv}{dt} \implies$ $$

$\dfrac{dv}{dt} = \dfrac{F * (1 - \dfrac{v^2}{c^2})^{\dfrac{3}{2}}}{m_{0}}$ $$

Using $v = 5 \: m/s, F = 1 \: N,$ and $m_{0} = 1 \: Kg$ with $c = 10 \: m/s \implies$ $$

$\dfrac{dv}{dt} = \dfrac{3 \sqrt{3}}{8} = .65 \: m/s^2$ to two decimal places. $$