Question 152 of 210

A block of mass $2kg$ moving on a horizontal surface with speed of $4{ms}^{-1}$ enters a rough surface ranging from $x=0.5m$ to $x=1.5m$. The retarding force in this range of rough surface is related to distance by $F=-kx$ where $k=12N{m}^{-1}$. The speed of the block as it just crosses the rough surface will be

A$2m{s}^{-1}$B$2.5m{s}^{-1}$C$1.5m{s}^{-1}$Dzero

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Answer: (A) $2m{s}^{-1}$

Acceleration of the block $a=\dfrac{f}{m}=-\dfrac{12x}{2}=-6x$ (Here the block is retarding)
Now, using, $a=v\dfrac{dv}{dx}$
$\Rightarrow vdv=adx$
On integrating, we get
${\int }_{4}^{v}vdv=-6{\int }_{0.5}^{1.5}xdx$
$\Rightarrow \dfrac{{v}^{2}-{4}^{2}}{2}=-6\left(\dfrac{{\left(1.5\right)}^{2}-{\left(0.5\right)}^{2}}{2}\right)$
$\Rightarrow {v}^{2}=16-12\Rightarrow v=2m{s}^{-1}$