Question

Two blocks A and B of equal masses are release from an inclined plane of inclination ${45}^o$ 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. Calculate when their front faces will come in a line. (Take $g=10m/s^2)$

Answer 1

Ans $\to$ coefficients of fraction between plane and block $A=0.2$

coefficients of fraction between plane and block $B=0.3$

$N_A=mg\cos \left({45}^o\right)$

$N_A=\frac{10m}{\sqrt{2}}\Rightarrow N=5\sqrt{2}m$

Frictional force acting on block $A=F_{RA}=N_A\times 0.2$

$\Rightarrow \frac{10m}{\sqrt{2}}\times \frac{2}{10}\Rightarrow F_{RA}=\sqrt{2}m$

Frictional force acting on block $B=F_{RB}$

$\Rightarrow \frac{10m}{\sqrt{2}}\times \frac{3}{10}\Rightarrow F_{RB}=\frac{3m}{\sqrt{2}}$

acceleration of block $A$ & $B$ clown the plane:-

$a=g\sin {45}^o\Rightarrow \left(10/\sqrt{2}\right)$

$a=5\sqrt{2}$

$a_r\to$ resultant acceleration := $m{a}_r=ma-F_{R}m{a}_r=m\left(5\sqrt{2}\right)=\frac{3m}{\sqrt{2}}$

$m{a}_r=ma-\sqrt{2}m{a}_r=5\sqrt{2}-3/\sqrt{2}$

$a_r=4\sqrt{2}m/s^2{a}_r=7/\sqrt{2}m/s^2$

$\ast$ Distance travelled by $A=5m$

$\ast$ Distance travelled by $B=(5-\sqrt{2})m$

for $A$:-

$s=\frac{1}{2}a{t}^1$

$s=\frac{1}{2}\left(4\sqrt{2}\right)t^2$

$t^2=\left(s/2\sqrt{2}\right)$

for $B$:-

$(s-\sqrt{2})=\frac{1}{2}\times \frac{7}{\sqrt{2}}{t}^2$

$(s-\sqrt{2})=\frac{7}{2\sqrt{2}}{t}^2$

put $t^2=\left(s/2\sqrt{2}\right)$

$\Rightarrow s=\sqrt{2}=\frac{7}{2\sqrt{2}}\left(\frac{s}{2\sqrt{2}}\right)$

$\Rightarrow s=8\sqrt{2}$ or $11.31m$

then $\to t^2=\frac{8\sqrt{2}}{2\sqrt{2}}\Rightarrow t=2s\to$ solution