Question

For the situation shown in figure, match the following table.


$\begin{array}{cc}\text{Table -1}& \text{Table -2}\\ \text{(A)Absolute acceleration of}& \text{(P)}11{\text{m/s}}^2\\ \text{1 kg block}& \\ \text{(B)Absolute acceleration of}& \text{(Q)}6{\text{m/s}}^2\\ \text{2 kg block}& \\ \text{(C)Relative acceleration between}& \text{(R)}17{\text{m/s}}^2\\ \text{the two blocks}& \\ & \text{(S) None}\end{array}$

• A. $\text{A}\to \text{S},\text{B}\to \text{P},\text{C}\to \text{S}$
• B. $\text{A}\to \text{S},\text{B}\to \text{S},\text{C}\to \text{S}$
• C. $\text{A}\to \text{P},\text{B}\to \text{P},\text{C}\to \text{P}$
• D. $\text{A}\to \text{P},\text{B}\to \text{S},\text{C}\to \text{P}$

Answer 1

The correct option is A $\text{A}\to \text{S},\text{B}\to \text{P},\text{C}\to \text{S}$

FBD of $1\text{kg}$ block:


From FBD, we get
$R_1=10\text{N}$
$f_1={\mu }_1{R}_1=0.4\times 10=4\text{N}$
[friction is kinetic]

FBD of $2\text{kg}$ block:


From FBD, we get

$R_2=R_1+20=30\text{N}$
$F={\mu }_2{R}_2=0.6\times 30=18\text{N}$

$\therefore$ Acceleration of $1\text{kg}$ block $\left(a_1\right)=\frac{f_1}{m_1}=\frac{4}{1}=4{\text{m/s}}^2\left(\text{leftwards}\right)$
Acceleration of $2\text{kg}$ block $\left(a_2\right)=\frac{F+f_1}{m_2}=\frac{18+4}{2}=11{\text{m/s}}^2\left(\text{rightwards}\right)$

Relative acceleration between the blocks $\left(a_{rel}\right)=|a_2-a_1|=|11+4|=15{\text{m/s}}^2$
Hence, option (a) gives the correct matching.