A force acts on a $3$ g particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where $x$ is in meters and $t$ is in seconds. The work done during the first $4$ second is :

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Answer 1

We have,
mass, $m = 3 g = 0.003 k g$
$x = 3 t - 4 t^{2} + t^{3}$

Now,
$v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2} \Rightarrow d x = (3 - 8 t + 3 t^{2}) d t$

$\Rightarrow a = \frac{d v}{d t} = 0 - 8 + 6 t$

Now,
$d w = F d x$

$\Rightarrow d w = (m a) d x$

$\Rightarrow d w = (0.003) (- 8 + 6 t) (3 - 8 t + 3 t^{2}) d t$

$\Rightarrow d w = (0.003) (18 t^{3} - 72 t^{2} + 82 t - 24) d t$

$\Rightarrow w = (0.003) \int_{0}^{4} (18 t^{3} - 72 t^{2} + 82 t - 24) d t$

$\Rightarrow w = 0.003 \times 176 = 0.528 J$

$\Rightarrow w = 530 m J$

Answer 2

We have,

mass,

m = 3 g = 0.003 k g

x = 3 t - 4 t^{2} + t^{3}

Now,

v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2} \Rightarrow d x = (3 - 8 t + 3 t^{2}) d t

\Rightarrow a = \frac{d v}{d t} = 0 - 8 + 6 t

Now,

d w = F d x

\Rightarrow d w = (m a) d x

\Rightarrow d w = (0.003) (- 8 + 6 t) (3 - 8 t + 3 t^{2}) d t

\Rightarrow d w = (0.003) (18 t^{3} - 72 t^{2} + 82 t - 24) d t

\Rightarrow w = (0.003) \int_{0}^{4} (18 t^{3} - 72 t^{2} + 82 t - 24) d t

\Rightarrow w = 0.003 \times 176 = 0.528 J

\Rightarrow w = 530 m J