The coefficient of static friction between the two blocks shown in figure is `mu` and the table is smooth. What maximum horizontal forced F can be applied to he block of mass M so that the block move together?

Options

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

Answer

D

Explanation

Here, `m_(A)=(m)/(2), m_(B)=m`
`mu_(A)=0.2, mu_(B)=0.1`
Let the blocks are moving with common acceleration a. THEN, `A=(mu_(A)m_(A)g)/(m_(A))=mu_(A)g=0.2g`
and `f-mug(m_(B)+m_(A))g=(m_(B)+m_(A))a`
`F=(m_(B)+m_(A))+mu_(B)(m_(B)+m_(A))g`
`(m+(m)/(g))(0.2g)+(0.1)(m+(m)/(2))g`
`((3)/(2)m)(0.2g)+((3)/(2)m)(0.1)g=(0.9)/(2)mg=0.45g`