A block of mass $5kg$ is connected to a vertical spring, as a result, the spring elongates by $4cm$. If another block of mass $2kg$ is attached to the block of mass $5kg$, determine the further elongation in spring.

Step 1: Given data

Mass of block connected to a vertical spring is $5kg$.

Elongation of the spring $=4cm=0.04m$.

Mass of the block attached to the $5kg$ block is $2kg$.

Step 2: To determine the spring constant of the given spring

Now, the weight force of the $5kg$ block $=mg=50N$

Since we know that,

$\Rightarrow kx=mg$

Where $k$ is the spring constant.

And $x$ is the elongation in the spring.

On putting the values we can write,

$\Rightarrow k(0.04)=50$

$\Rightarrow k=\frac{50}{0.04}=1250N{m}^{-1}$

Hence the spring constant of the given spring is $1250N{m}^{-1}$.

Step 3: To determine the further elongation in spring after $\mathbf{2}\mathbf{}\mathit{k}\mathit{g}\mathbf{}$block is added

Now. when another $2kg$ block is added,

Let the further elongation in the spring be $x_0$,

Since we know that,

$\Rightarrow kx=mg$

Where $k$ is the spring constant.

And $x$ is the elongation in the spring.

On putting the required data in equation, we get,

$\Rightarrow k\times (0.04+x_0)=(5+2)g$

$\Rightarrow 1250\left(0.04+x_0\right)=70$

On further solving the equation for $x_0$, then

$\Rightarrow \left(0.04+x_0\right)=\frac{7}{125}$

$\Rightarrow x_0=0.016m$

Since the answer is in centimeters we can convert it,

$\Rightarrow x_0=1.6cm$

Hence the further elongation in spring is $\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{}\mathit{c}\mathit{m}$.