A force acts on a 3, gm particle in such a wa | vedclass

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Question

$x=3 t-4 t^{2}+t^{3}$

$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=3-8 \mathrm{t}+3 \mathrm{t}^{2}$

${\mathrm{t}=0,}  {\mathrm{v}_{1}=3 \mat

Vedclass · Accepted Answer

$528$

Vedclass · Answer

$x=3 t-4 t^{2}+t^{3}$

$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=3-8 \mathrm{t}+3 \mathrm{t}^{2}$

${\mathrm{t}=0,}  {\mathrm{v}_{1}=3 \mathrm{m} / \mathrm{s}} $

$ {\mathrm{t}=4,}  {\mathrm{v}_{2}=3-8(4)+3(4)^{2}}$

$\mathrm{v}_{2} \Rightarrow 19 \mathrm{m} / \mathrm{s}$

$\mathrm{w}_{1 ightarrow 2}=\mathrm{k}_{2}-\mathrm{k}_{1}$

${=\frac{1}{2} \mathrm{m}\left(\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}ight)} $

${=\frac{1}{2} 	imes 3 	imes 10^{-3}\left(19^{2}-3^{2}ight)} $

${=528 \mathrm{mJ}}$

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 ................. $\mathrm{mJ}$

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