Question

Two stationary blocks $A$ and $B$ of equal masses are released from an inclined plane of inclination $60^\circ $ at $t=0,$ such that block $A$ is $2.4 \,m$ behind block $B$ . The coefficient of kinetic friction between the block $A$ and the inclined plane is $0.4$ while it is $0.6$ for block $B$ . If both the blocks meet after $A$ has travelled a distance $5.95\alpha $ ( in $m$ ) down the plane then find value of $\alpha $ .(Take $g = 10 \,m / s^{2} $ and $ \sqrt{3} = 1 . 73$

Answer 1

Acceleration of A down the plane,
$a_{A}=gsin60^\circ -\mu _{A}gcos60^\circ $
$=10\times \frac{\sqrt{3}}{2}-0.4\times 10\times \frac{1}{2}=5\sqrt{3}-2$
$\therefore a_{A}=6.65\,m/s^{2}$
Acceleration of $B$ down the plane,
$a_{B}=gsin60^\circ -\mu _{B}gcos60^\circ $
$=10\times \frac{\sqrt{3}}{2}-0.6\times 10\times \frac{1}{2}=5\sqrt{3}-3$
$\therefore a_{B}=5.65\,m/s^{2}$
The front face of $A$ and $B$ will come in line when, $s_{A}=s_{B}+2.4$
$\therefore \frac{1}{2}a_{A}t^{2}=\frac{1}{2}a_{B}t^{2}+2.4$
$\therefore \frac{1}{2}\times 6.65\times t^{2}=\frac{1}{2}\times 5.65\times t^{2}+2.4$
$\therefore 3.33t^{2}=2.83t^{2}+2.4$
$\therefore 0.5t^{2}=2.4$
$\therefore t^{2}=4.8$
$\therefore t=2.19\,s$
Now, Now, $s_A=\frac{1}{2} a_A t^2=\frac{1}{2} \times 6.65 \times(2.19)^2$
$\therefore s_{\Lambda}=15.95\,m=15.95\times 1\,m$