Question

In a tank of horizontal cross-sectional area $1m^2$, a spring with force constant $2000N{m}^{-1}$ is fixed in vertical position upto the height of the water as shown in figure I. A block of mass $180kg$ is gently placed over the spring and it attains the equilibrium position as shown in figure II. If base area of the block is $0.2m^2$ and height $60cm$, then find compression (in $cm$ ) in the spring in equilibrium position.
(Take $g=10m/s^2,{\rho }_w=1000kg/m^3$ )

Answer 1

Answer: 40

As block goes down by distance $x$, water comes up by distance $y$. As both are measured from initial level of water, compression in the spring is $x$ but the block is in depth $(x+y)$ in water.
Applying conservation of volume

$0.2\times x=\left(1m^2-0.2m^2\right)\times y$
$x=4y\Rightarrow y=\frac{x}{4}$
Thus, total depth of block in water $=\frac{x}{4}+x=\frac{5x}{4}$
Free body diagram of the block in equilibrium:
$F_b=(0.2)(5x/4)(1000)(10)$
For equilibrium,
$mg=kx+F_B$

$\Rightarrow 1800=2000x+(0.2)\left(\frac{5x}{4}\right)(1000)(10)$
$\Rightarrow 18=20x+25x$
$\Rightarrow x=\frac{18}{45}m=40cm$