Question

A block of mass m = 1 kg moving on a horizontal surface with speed $v_i=2m{s}^{-1}$ enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding force $F_r$ on the block in this range is inversely proportional to x over this range.
$F_r=-\frac{k}{x}$ for $0.1<x<2.01$. $F_r=0$ for $x<0.1m$ and $x>2.01m$. What is the final speed$\left(\frac{m}{s}\right)$ of the block as it crosses the path? (Given: log 20.1 = 1.30)

Answer 1

$Here,m=1kg,v_i=2m{s}^{-1};k=0.5J$

Initial K.E. $K_i=\frac{1}{2}m{v}_i^2=\frac{1}{2}\times 1{\left(2\right)}^2=2J$

Work done against friction

W = ${\int }_{0.1}^{2.01}{f}_{r}dx={\int }_{0.1}^{2.01}-\frac{k}{x}dx$

$=-0.5{\left[lo{g}_{e}x\right]}_{0.1}^{2.01}$

= $-0.5lo{g}_e\frac{2.01}{0.1}$

W =$-0.5\times 2.303lo{g}_{10}20.10$

W = $-0.5\times 2.303\times 1.303=-1.5J$

$\therefore FinalK.E.,K_f=K_i+W$

= 2.0 - 1.5 J = 0.5 J

$v_f=\sqrt{\frac{2k_f}{m}}=\sqrt{\frac{2\times 0.5}{1}}=1m{s}^{-1}$