Question

The masses of the blocks A and B are $0.5\text{kg}$ and $1\text{kg}$ respectively. These are arranged as shown in the figure and are connected by a massless string. The coefficient of friction between all contact surfaces is $0.4$. The force, necessary to move the block B with constant velocity, will be $(g=10{\text{m/s}}^2)$

• A. $5\text{N}$
• B. $10\text{N}$
• C. $15\text{N}$
• D. $20\text{N}$

Answer 1

The correct option is B $10\text{N}$

Taking $m_A=0.5\text{kg};m_B=1\text{kg}$
Force on block A ....(1)
$T=\mu m_{A}g$
Force acting on block B
$F=T+\mu m_{A}g+\mu (m_A+m_B)g|....\left(2\right)$
From (1) & (2)
$F=\mu m_{A}g+\mu m_{A}g+\mu m_{A}g+\mu m_{B}g$
$F=3\mu m_{A}g+\mu m_{B}g=\mu g(3m_A+m_B)$
$=0.4\times 10\times (3\times 0.5+1)=10\text{N}$