Question

A small block is freely sliding down from the top of a rough inclined plane whose angle of inclination is ${45}^0$. The block reaches bottom then it completes a smooth vertical circle. If the coefficient of friction is $0.5$, the ratio of minimum vertical height of inclined plane to radius of vertical circle is:

• A. $3:1$
• B. $5:1$
• C. $5:2$
• D. $7:3$

Answer 1

The correct option is B $5:1$

$\mu =0.5$.

$W_{o}byfriction=\mu Na$

$=\left(0.5\right)\left(\frac{mg}{\sqrt{2}}\right)\left(\sqrt{2H}\right)$

$=\frac{mgH}{2}$

minimum velocity required for completely velocity lively$=\sqrt{5gR}$ work energy therefore on work

$mgH=\frac{m{V}_o^2}{2}+\frac{mgH}{2}.$

$\frac{mgH}{2}=\frac{m}{2}{\left(\sqrt{5gR}\right)}^2$

$gH=5gR.$

$\frac{H}{R}=\frac{5}{1}$