Question

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

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

Answer 1

Answer: (C)

$\frac{1}{K}(\sqrt{{\mu }^2{m}^2{g}^2+mK{u}^2}-\mu mg)$

Using work-energy theorem,
$\frac{1}{2}m{u}^2=\mu mgS+\frac{1}{2}K{S}^2$
i.e. $S^2+\frac{2\mu mgS}{K}-\frac{m{u}^2}{K}=0$
Solving for S we get,
$S=\frac{1}{K}(\sqrt{{\mu }^2{m}^2{g}^2+mK{u}^2}-\mu mg)$