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A force acts on a $3\, gm$ particle in such a way that the position of the particle as a function of time is given by $x = 3t -4t^2 + t^3$ , where $x$ is in $meters$ and $t$ is in $seconds$ . The work done during the first $4\, second$ is ................. $\mathrm{mJ}$
$384$
$168$
$528$
$541$
Solution
$x=3 t-4 t^{2}+t^{3}$
$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=3-8 \mathrm{t}+3 \mathrm{t}^{2}$
${\mathrm{t}=0,} {\mathrm{v}_{1}=3 \mathrm{m} / \mathrm{s}} $
$ {\mathrm{t}=4,} {\mathrm{v}_{2}=3-8(4)+3(4)^{2}}$
$\mathrm{v}_{2} \Rightarrow 19 \mathrm{m} / \mathrm{s}$
$\mathrm{w}_{1 \rightarrow 2}=\mathrm{k}_{2}-\mathrm{k}_{1}$
${=\frac{1}{2} \mathrm{m}\left(\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}\right)} $
${=\frac{1}{2} \times 3 \times 10^{-3}\left(19^{2}-3^{2}\right)} $
${=528 \mathrm{mJ}}$
