Question

Two blocks $A$ and $B$ of equal masses are sliding down along straight parallel lines on an inclined plane of ${45}^{\circ }$. Their coefficients of kinetic friction are ${\mu }_A=0.2$ and ${\mu }_B=0.3$ respectively. At $t=0$, both the blocks are at rest and block $A$ is $\sqrt{2}$ meter behind block $B$. What is the time (in second) taken from the initial position where the front faces of the blocks come in line on the inclined plane as shown in figure. (Use $g=10m{s}^{-2})$.

Answer 1

Let acceleraton of block A and B is $a_A$ and $a_B$ along inclined plane.

$mg\sin {45}^{\circ }-f_A=m{a}_A$

$mg\sin {45}^{\circ }-{\mu }_{A}mg\cos {45}^{\circ }=m{a}_A$

$\frac{g}{\sqrt{2}}(1-{\mu }_A)=a_A$

$\frac{10}{\sqrt{2}}(1-0.2)=a_A$

$a_A=\frac{8}{\sqrt{2}}$

from fig. 2;

$mg\sin {45}^{\circ }-f_B=m{a}_B$

$mg\sin {45}^{\circ }-{\mu }_{B}mg\cos {45}^{\circ }=m{a}_B$

$\frac{g}{\sqrt{2}}(1-{\mu }_B)=a_B$

$a_B=\frac{7}{\sqrt{2}}$

Let acceleration of block with respect to block B is

$a_{AB}=a_A-a_B=\frac{1}{\sqrt{2}}m/s^2$

Relative displacement is $\sqrt{2}m$.

Time taken to cover is $t$.

$\sqrt{2}=\frac{1}{2}{a}_{AB}{t}^2$

$\sqrt{2}=\frac{1}{2}\times \frac{1}{\sqrt{2}}{t}^2$

$t^2=4$

$t=2s$