Question

A block of mass $m$ has initial velocity $u$ having direction towards '+x axis'. The block stops after covering a distance $S$ causing similar extension in the spring of constant $K$ holding it, $\mu$ is the kinetic friction between the block and the surface on which it was moving, the distance $S$ is given by:

• A. $\frac{1}{K}{\mu }^2{m}^2{g}^2$
• B. $\frac{1}{K}(mK{u}^2-{\mu }^2{m}^2{g}^2{)}^{1/2}$
• C. $\frac{1}{K}({\mu }^2{m}^2{g}^2+mK{\mu }^2+\mu mg{)}^{1/2}$
• D. $\frac{\mu mg+\sqrt{{\mu }^2{m}^2{g}^2+m{u}^{2}k}}{k}$

Answer 1

The correct option is D $\frac{\mu mg+\sqrt{{\mu }^2{m}^2{g}^2+m{u}^{2}k}}{k}$

By the work-energy theorem $\frac{1}{2}m{u}^2=\mu mgS+\frac{1}{2}k{S}^2$
i.e $S^2+\frac{2\mu mgS}{k}-\frac{m{u}^2}{k}=0$
$\Rightarrow \frac{\frac{-2\mu mg}{k}+\sqrt{\frac{4{\mu }^2{m}^2{g}^2}{k^2}+\frac{4m{u}^2}{k}}}{2}$
$=\frac{-\mu mg+\sqrt{{\mu }^2{m}^2{g}^2+m{u}^{2}k}}{k}$