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 $g=10 \, m \, s^{- 2}$ )

Answer 1

Acceleration of A down the plane,
a_A = g sin 45° -μ_A g cos 45^o
$= \left(\right. 10 \left.\right) \left(\right. \frac{1}{\sqrt{2}} \left.\right) - \left(\right. 0.2 \left.\right) \left(\right. 10 \left.\right) \left(\right. \frac{1}{\sqrt{2}} \left.\right)$
= 4√2m/s²
Similarly acceleration of B down the plane, a_B =g (sin 45°- μ_B cos 45°)
$= \left(\right. 10 \left.\right) \left(\right. \frac{1}{\sqrt{2}} \left.\right) - \left(\right. 0.3 \left.\right) \left(\right. 10 \left.\right) \left(\right. \frac{1}{\sqrt{2}} \left.\right) = 3.5 \sqrt{2} \frac{m}{s^{2}}$
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, $s_{A} = \frac{1}{2} a_{A} t^{2} = \frac{1}{2} \times 4 \sqrt{2} \times \left(\right. 2 \left.\right)^{2} = 8 \sqrt{2} m$
So, both the blocks will come in a line after A has travelled a distance 8√2 m down the plane.