Question

Force constant of a spring is $100\text{N/m}$. If both the blocks of mass $10\text{kg}$ and $12\text{kg}$ attached with spring and string are at rest. Then find the extension in spring. Take $g=10{\text{m/s}}^2$.

• A. $1.1\text{m}$
• B. $2.2\text{m}$
• C. $3.3\text{m}$
• D. $1.2\text{m}$

Answer 1

The correct option is B $2.2\text{m}$

When both blocks are in equilibrium,then vertical forces should be balanced on both blocks.

From the FBD of both blocks shown below


For $12\text{kg}$ block; Balancing the vertical forces we get,

$T-120=0$
$\therefore T=120\text{N}$

Let the extension in spring be $x\text{m}$ at equilibrium, the spring force will be $kx$ hence balancing the forces in vertical direction we get,

$\Rightarrow kx-100-T=0$
$\therefore 100x-100-120=0$
i.e $x=\frac{220}{100}=2.2\text{m}$

Hence option B is the correct answer