A force acts on a 3.0 g particle in such a wa | vedclass

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Question

$x=3 t-4 t^{2}+t^{3}$

$\frac{d x}{d t}=3-8 t+3 t^{2}$ and $a=\frac{d^{2} x}{d t^{2}}=-8+6 t$

Now, $\quad W=\int F d x=\int m a d x=\int m a \fra

Vedclass · Accepted Answer

$528$

Vedclass · Answer

$x=3 t-4 t^{2}+t^{3}$

$\frac{d x}{d t}=3-8 t+3 t^{2}$ and $a=\frac{d^{2} x}{d t^{2}}=-8+6 t$

Now, $\quad W=\int F d x=\int m a d x=\int m a \frac{d x}{d t} d t$

$=\int_{0}^{4} \frac{3}{1000} 	imes(-8+6 t)\left(3-8 t+3 t^{2}ight) d t$

On integrating, we get; $W=528 \mathrm{mJ}$

A force acts on a $3.0\ g$ 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 metres and $t$ is in seconds. The work done during the first $4\ s$ is ................. $\mathrm{mJ}$

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