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 $2.4m$ 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 $g=10m/s^2$ and $\sqrt{3}=1.73$

Answer 1

Answer: 1

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