Question

The system is released from rest with both the springs in unstretched positions. Mass of each block is $1\text{kg}$ and force constant of each spring is $10\text{N/m}$. Extension of horizontal spring in equilibrium is:

• A. $0.2\text{m}$
• B. $0.4\text{m}$
• C. $0.6\text{m}$
• D. $0.8\text{m}$

Answer 1

The correct option is B $0.4\text{m}$

$K=10\text{N/m}$
For lower block :
$2T+F_s=Mg$
$2T+Kx/2=Mg$ ... (1)
For block placed on horizontal level :
$Kx=T$ ... (2)
Putting (1) in (2)
$T=\frac{Mg-Kx/2}{2}$
$Kx=\frac{2Mg-Kx}{4}$
$5Kx=2\times 1\times 10$
$5\times 10x=2\times 10$
$x=2/5=0.4\text{m}$