Question

The coefficient of static friction, ${\mu }_s$, between block $A$ of mass 2 kg and the table as shown in the figure is $\mathrm{0.2.}$ What would be the maximum mass value of block $B$ so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless. $(g=10{\text{m/s}}^2)$

• A. 2.0 kg
• B. 4.0 kg
• C. 0.2 kg
• D. 0.4 kg

Answer 1

The correct option is D 0.4 kg

Free body diagrams of two masses are


We get equations
$T-f=m_{A}aandf=\mu N=\mu m_{A}g$
$\Rightarrow T=m_{A}a+\mu m_{A}gandT=m_{B}g(fora=0)$
$\therefore \mu N=m_{B}g\Rightarrow m_B=\mu m_A=0.2\times 2=0.4\text{kg}.$
Hence, the correct answer is option (d).