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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
$x=3 t-4 t^{2}+t^{3}$
$\mathrm{v}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \mathrm{t}}=3-| \mathrm{8t}+3 \mathrm{t}^{2}$
$\mathrm{t}=0, \quad \mathrm{v}_{1}=3 \mathrm{m} / \mathrm{s}$
$\mathrm{t}=4, \quad \mathrm{v}_{2}=3-8(4)+3(4)^{2}$
$v_{2} \Rightarrow 19 \mathrm{m} / \mathrm{s}$
$\mathrm{w}_{1 \rightarrow 2}=\mathrm{k}_{2}-\mathrm{k}_{1}$
$=\frac{1}{2} \mathrm{m}\left(\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}\right)$
$=\frac{1}{2} \times 3 \times 10^{-3}\left(19^{2}-3^{2}\right)$
$=528 \mathrm{mJ}$
