Question

The blocks A & B have masses of $2$ & $6$ kg respectively are kept on a rough horizontal surface of $\mu =0.2$. If a horizontal force of $6$ N pushes them. Calculate the acceleration of the system.

Answer 1

Now, the maximum frictional force applied by the surface on the two blocks is as follow:

The maximum frictional force on the block A is,

$F_{fa}=\mu \left(N_A\right)$

$F_{fa}=\mu (m_A\times g)$

$F_{fa}=0.2(2kg\times 9.8m/s^2)$

$F_{fa}=3.92N$

Here, the maximum frictional force applied by the surface on the block A is $F_{fa}$

The maximum frictional force on the block B is,

$F_{fb}=\mu \left(N_B\right)$

$F_{fb}=\mu (m_B\times g)$

$F_{fb}=0.2(6kg\times 9.8m/s^2)$

$F_{fb}=11.76N$

Here, the maximum frictional force applied by the surface on the block B is $F_{fb}$

so, now the total maximum frictional force that can be applied by the surface on the two block is $F_f=F_{fa}+F_{fb}$

$F_f=3.92N+11.76N$

$F_f=15.68N$

But, here the external force on the blocks is only $6N$ hence, the applied force is not enough to move the blocks because, the frictional force will neutralize the force. So, the acceleration of the system is zero.

This is same for the both the figures.