4-1.Newton's Laws of Motion
hard

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

A$0$
B$1.5$
C$2.0$
D$2.5$
(JEE MAIN-2022)

Solution

$a=\frac{- kx }{2}=\frac{-12 x }{2}=-6 x$
$\frac{ v d v }{ dx }=-6 x$
$\int \limits_{4}^{*} vdv =-\int \limits_{\frac{1}{2}}^{ s / 2} 6 xdx$
$\frac{ v ^{2}-4^{2}}{2}=-\frac{6}{2}\left[\left(\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}\right]$
$v ^{2}-16=-6\left(\frac{9}{4}-\frac{1}{4}\right)$
$v ^{2}=16-6 \times 2=4$
$V =2\,m / s$
Standard 11
Physics

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