(A) To complete a vertical circular path,the minimum velocity $v$ at the bottom of the circle must be $v = \sqrt{5gR}$.Given $R = 2\ m$ and $g = 10\ m

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(A) To complete a vertical circular path,the minimum velocity $v$ at the bottom of the circle must be $v = \sqrt{5gR}$.Given $R = 2\ m$ and $g = 10\ m/s^2$,the required velocity is $v = \sqrt{5 \times 10 \times 2} = \sqrt{100} = 10\ m/s$.On the rough surface,the initial velocity $u = 20\ m/s$,final velocity $v = 10\ m/s$,and acceleration $a = -\mu g = -0.5 \times 10 = -5\ m/s^2$.Using the equation of motion $v^2 - u^2 = 2ax$:$10^2 - 20^2 = 2 \times (-5) \times x$$100 - 400 = -10x$$-300 = -10x$$x = 30\ m$.

Class 11 Physics Work, Energy, Power and Collision Vertical Circular Motion

Difficult

$A$ block of mass $m$ is given a velocity $20\ m/s$ on a horizontal rough surface of length $x\ m$. After the rough surface,there is a vertical circular path of radius $R = 2\ m$ which is frictionless. The block enters the circular path with the minimum velocity required to complete the path. Find the length of the rough surface $x$ if the coefficient of friction is $\mu = 0.5$.

A
$30$
B
$20$
C
$10$
D
$16$

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