Question

A block of mass $2\text{kg}$ is resting over another block of mass $6\text{kg}$. $2\text{kg}$ block is connected to one end of a string fixed to a vertical wall as shown. If the coefficient of friciton between the blocks is $0.4$, the force required to pull out the $6\text{kg}$ block with an acceleration of $1.5{\text{m/s}}^2$ will be $(g=10{\text{ms}}^{-2})$

• A. $17\text{N}$
• B. $9\text{N}$
• C. $8\text{N}$
• D. $7\text{N}$

Answer 1

The correct option is A $17\text{N}$


As $2\text{kg}$ block is connected to wall hence always at rest.
Making F.B.D
For $2\text{kg}$ block
(1) vertical balance $=N_2=2\text{g}=20\text{N}$
$f_r=\mu N_2$
$f_r=0.4\times 20=8\text{N}$
Now for $6\text{kg}$ block
$f-f_r=6a$
$=f-8=6\times 1.5$ (given)
$=f=9+8$
$f=17\text{N}$