Question

In the arrangement shown in figure $m_A=m_g=2kg$. Block B is resting on a smooth horizontal surface, while friction coefficient between blocks A and B is $\mu =0.5$. The maximum horizontal force F can be applied so that block A does not slip over not slip over the block B is $(g=10m{s}^{-2})$

• A. 25 N
• B. 40 N
• C. 30 N
• D. 20 N

Answer 1

The correct option is D 20 N

Let maximum friction is acting $f=0.5\times 2\times 10=10N$

and both the block as a acceleration a

in A block

F-f=2a ...eq(1)

in B block

f=2a... eq(2)

and f=10N

from eq(2)

$a=5m{s}^{-2}$

from eq(1)

$F=2\times 5+10=20N$

Hence the D option is correct.