Q: A block starts moving up an inclined plane of inclination 30° with an initial velocity vo . It comes back to its initial position with velocity vo/2 . The value of coefficient of kinetic friction between the block and inclined plane is close to I/1000 . The nearest integer to I is …….
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Sol: 
When block is moving up the plane ,
Retardation a1 = gsin30° + μ g cos30°
$\displaystyle a_1 = \frac{g}{2} + \frac{\mu g \sqrt{3}}{2}$
$\displaystyle v^2 = u^2 + 2 a s$
$\displaystyle 0 = v_o^2 – 2 (\frac{g}{2} + \frac{\mu g \sqrt{3}}{2}) s$
$\displaystyle v_o^2 = g(1 + \sqrt{3}\mu)$ …(i)
When block is moving down the plane ,
Retardation , $\displaystyle a_2 = gsin30° – μ g cos30° $
$\displaystyle a_2 = \frac{g}{2} – \frac{\mu g \sqrt{3}}{2}$
writing equation for down the plane ,
$\displaystyle (\frac{v_o}{2})^2 = 2 (\frac{g}{2} – \frac{\mu g \sqrt{3}}{2}) s $
$\displaystyle \frac{v_o^2}{4} = 2 (\frac{g}{2} – \frac{\mu g \sqrt{3}}{2}) s $
$\displaystyle v_o^2 = 4 g(1 – \sqrt{3}\mu)$ ….(ii)
From (i) & (ii)
$\displaystyle g(1 + \sqrt{3}\mu) = 4 g(1 – \sqrt{3}\mu)$
$\displaystyle (1 + \sqrt{3}\mu) = 4(1 – \sqrt{3}\mu)$
$\displaystyle 5 \sqrt{3}\mu = 3 $
$\displaystyle \mu = \frac{3}{5 \sqrt{3}}= \frac{\sqrt{3}}{5} $
$\displaystyle \mu = \frac{1.732}{5} \times \frac{200}{200}$
$\displaystyle \mu = \frac{346.2}{1000} $
Therefore I = 346
