Question

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:

• A. Zero
• B. $1.5 \, ms ^{-1}$
• C. $2.0\, ms ^{-1}$
• D. $2.5 \, ms ^{-1}$

Answer 1

Answer: (C)

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