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Work Done

The position ...

Question

The position of a particle of mass $4 g$ , acted upon by a constant force is given by $x = 4 t^{2} + t$ , where $x$ is in metre and $t$ in second. The work done during the first $2 s$ is

128 m J

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512 m J

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576 m J

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144 m J

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288 m J

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Solution

The correct option is D $576 m J$
The position of a particle
$x = u t^{2} + t$
$v = \frac{d x}{d t} = 8 t$
$a = \frac{d^{2} x}{d t} = 8$
The distance in first $2 s$
$x = 4 t^{2} + t = 4 \times {(2)}^{2} + 2$
$x = 18 m$
The work done during the first $2 s$
$W = F \cdot x$
$= m \cdot a \cdot x$
$= 4 \times 10^{- 3} \times 8 \times 18$
$= 576 \times 10^{- 3} J$
$W = 576 m J$

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The position of a particle of mass 4g, acted upon by a constant force is given by x=4t2+t, where x is in metre and t in second. The work done during the first 2s is

The correct option is D 576mJThe position of a particlex=ut2+tv=dxdt=8ta=d2xdt=8The distance in first 2sx=4t2+t=4×(2)2+2x=18mThe work done during the first 2sW=F⋅x =m⋅a⋅x =4×10−3×8×18 =576×10−3JW=576mJ

The position of a particle of mass $4 g$ , acted upon by a constant force is given by $x = 4 t^{2} + t$ , where $x$ is in metre and $t$ in second. The work done during the first $2 s$ is