Question

The system is released from rest with both the springs in unstretched position. Mass of each block is $5\text{kg}$ and force constant of each spring is $10\text{N/m}$. Assume pulley and strings are massless and all contacts are smooth. Take $g=10{\text{m/s}}^2$

• A. Extension in horizontal spring at equilibrium is $2\text{m}$
• B. Compression in vertical spring at equilibrium is $1\text{m}$
• C. Compression in vertical spring at equilibrium is $2\text{m}$
• D. Extension in horizontal spring at equilibrium is $3\text{m}$

Answer 1

Answer: (A), (B)

Compression in vertical spring at equilibrium is $1\text{m}$

If $B$ moves downwards by distance $x$, pulley attached by string to block $B$ will also move $x$ downwards
So, block $A$ will move by distance $2x$ rightwards.
$\therefore$ Compression in vertical spring$=x$
$\&$ extension in horizontal spring$=2x$

If tension in string connected to block $A$ is $T$, the tension in string connnecting block $B$ and pulley is $2T$

F.B.D of blocks (at equilibrium $a=0$)


For block $A$, equilibrium condition in horizontal:
$T=k\left(2x\right)...\left(i\right)$
For block $B$, equilibrium condition in vertical:
$2T+kx=mg...\left(ii\right)$
From Eq $\left(i\right)\&\left(ii\right)$,
$\Rightarrow 2\left(2kx\right)+kx=mg$
Or, $5kx=mg$
$\therefore x=\frac{mg}{5k}=\frac{5\times 10}{5\times 10}=1\text{m}$
Hence, compression in vertical spring is $x=1\text{m}$, extension in horizontal spring is $2x$, i.e $2\text{m}$.