Physics - W.E.P. Question with Solution | TestHub

$\because$According to the question,

$\frac{1}{2}m{v}_f^2=\frac{1}{2}\times \frac{1}{2}m{v}_i^2$

$\left(v_i, v_f=\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{b}\mathrm{o}\mathrm{d}\mathrm{y}\right)$

or$v_f^2=\frac{v_i^2}{2}$or$v_f=\frac{10}{\sqrt{2}}$

Given,$F=-kv$

or$ma= -kv\Rightarrow \frac{mdv}{dt}=-kv$

$\Rightarrow \underset{10}{\overset{\frac{10}{\sqrt{2}}}{\int }}\frac{1}{v}dv= -\underset{0}{\overset{10}{\int }}kdt \left(\because m=1kg\right)$

$\Rightarrow {{\left(ln\mathrm{ }v\right)}_{10}}^{\frac{10}{\sqrt{2}}}=-k\left(10\right)$

$\Rightarrow ln\frac{\mathrm{ 10}}{\sqrt{2}}-ln\mathrm{ 10}=-k\left(10\right)$

$\Rightarrow k=\frac{1}{10}ln\left(\frac{10}{\frac{10}{\sqrt{2}}}\right)=\frac{1}{10}ln\sqrt{2}=\frac{\ln 2}{20}$