Question: In a tank of horizontal cross-sectional area 1m<sup>2</sup>, a spring with force constant 2000 Nm<su...

Question

In a tank of horizontal cross-sectional area 1m², a spring with force constant 2000 Nm^-1 is fixed in vertical position upto the height of the water as shown in figure I. A block of mass 180 kg is gently placed over the spring and it attains the equilibrium position as shown in figure II. If base area of the block is 0.2 m² and height 60 cm, then compression in the spring in equilibrium position is (take g = 10 m/s²):

Answer 1

Solution

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 × x

= (1m² –0.2m²).y

x = 4y ⇒ y = $\frac{x}{4}$

Thus total depth of block in water = $\frac{x}{4}$ + x = $\frac{5x}{4}$

Free body diagram in equilibrium :

1800 = 2000.x+(0.2) $\left( \frac{5x}{4} \right)$ (1000) (10)

⇒ 18 = 20x + 25x

⇒ x = $\frac{18}{45}$ m=40 cm.