Question

Block A of mass $M$ in the system shown in the figure slides down the incline at a constant speed. The coefficient of friction between block A and the surface is $\frac{1}{3\sqrt{3}}$. The mass of block $B$ is

• A. $M/2$
• B. $M/3$
• C. $2M/3$
• D. $M/\sqrt{3}$

Answer 1

The correct option is B $M/3$

Given,

Block A moving with constant velocity, mean net force on block A is zero

Force equilibrium on block A

$F-F_r-T=0$

Where,

$F\left(forceduetogaravitycomponent\right)=Mg\sin \theta$

$F_r\left(Frictionforce\right)=\mu Mg\cos \theta$

$T\left(tensionincable\right)=weigthofblockB=M_{B}g$

$Mg\sin {30}^o-\mu Mg\cos {30}^o-M_{B}g=0$

$M\frac{1}{2}-\frac{1}{3\sqrt{3}}M\frac{\sqrt{3}}{2}=M_B$

$M_B=\frac{M}{3}$

Mass of block B is $\frac{M}{3}$