Question

Two blocks $A$ and $B$ of equal masses are released from an inclined plane of inclination $45^{\circ }$ at $t=0$ . Both the blocks are initially at rest. The coefficient of kinetic friction between the block $A$ and the inclined plane is $0.2$ while it is $0.3$ for block $B$ . Initially the block $A$ is $\sqrt{2}m$ behind the block $B$ . At what time in seconds will their front faces come in a line, (Take $g=10m{s}^{-2}$ )

Answer 1

Answer: 2

Acceleration of A down the plane,
a_A = g sin 45° -μ_A g cos 45^o
$=(10)(\frac{1}{\sqrt{2}})-(0.2)(10)(\frac{1}{\sqrt{2}})$
= 4√2m/s²
Similarly acceleration of B down the plane, a_B =g (sin 45°- μ_B cos 45°)
$=(10)(\frac{1}{\sqrt{2}})-(0.3)(10)(\frac{1}{\sqrt{2}})=3.5\sqrt{2}\frac{m}{s^2}$
The front face of A and B will come in a line when,
s_a = s_B + √2
or $\frac{1}{2}{u}_A{t}^2=\frac{1}{2}{a}_B{t}^2+\sqrt{2}$
$\frac{1}{2}\times 4\sqrt{2}\times t^2=\frac{1}{2}\times 3.5\sqrt{2}\times t^2+\sqrt{2}$
Solving this equation, we get t = 2 s
Additional information, $s_A=\frac{1}{2}{a}_A{t}^2=\frac{1}{2}\times 4\sqrt{2}\times (2{)}^2=8\sqrt{2}m$
So, both the blocks will come in a line after A has travelled a distance 8√2 m down the plane.