Question

In figure, the coefficient of friction between the floor and the block B is 0.1. The coefficent of friction between the blocks B and A is 0.2. The mass of A is m/2 and B is m. What is the maximum horizontal force F can be applied to the block B so that two blocks move together?

• A. 0.15 mg
• B. 0.05 mg
• C. 0.1 mg
• D. 0.45 mg

Answer 1

The correct option is D 0.45 mg

Here, $m_A=\frac{m}{2},m_B=m$

${\mu }_A=0.2,{\mu }_B=0.1$

Let both the blocks are moving with common acceleration a. Then,

From FBD of the block.

$a=\frac{{\mu }_A{m}_{A}g}{m_A}={\mu }_{A}g=0.2g$

and $F-{\mu }_B(m_B+m_A)g=(m_B+m_A)a$

$F=(m_B+m_A)g+\mu (m_B+m_A)g$

=$(m+\frac{m}{2})\left(0.2g\right)+\left(0.1\right)(m+\frac{m}{2})g$

=$\left(\frac{3}{2}m\right)\left(0.2g\right)+\left(\frac{3}{2}m\right)\left(0.1g\right)=\frac{0.9}{2}mg=0.45mg$