Physics - NLM Question with Solution | TestHub

Distance travelled by the body in$n^{\mathrm{t}\mathrm{h}}$second is given by

$S_n=u+\frac{a}{2}(2n-1)$

$5=u+\frac{a}{2}\left(2\times 1-1\right)$

$5=u+\frac{a}{2} \dots ..\left(\mathrm{i}\right)$

$2=u+\frac{a}{2}\left(2\times 3-1\right)$

$2=u+\frac{5}{2}a\dots .\left(\mathrm{i}\mathrm{i}\right)$

Solving Eqs. (i) and (ii), we get

$a=-\frac{6}{\mathrm{4 }}{\mathrm{m }\mathrm{s}}^{-2}$

$ie, \mathrm{b}\mathrm{o}\mathrm{d}\mathrm{y}\mathrm{ }$is decelerating

$\mathrm{G}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n},\mathrm{ }\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}=\mathrm{4 }\mathrm{k}\mathrm{g}$

$\therefore F=m\times a=4\times \frac{6}{4}=6\mathrm{ }\mathrm{N}$