Question

A block of mass $2kg$ moving on a horizontal surface with speed of $4m{s}^{-1}$ 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. Zero
• B. $1.5m{s}^{-1}$
• C. $2.0m{s}^{-1}$
• D. $2.5m{s}^{-1}$

Answer 1

Answer: (C)

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