Question

A block of density $2000{\text{kg/m}}^3$ and mass $10\text{kg}$ is suspended by a spring of stiffness $100$ $\text{N/m}$. The other end of the spring is attached to a fixed support. The block is completely submerged in a liquid of density $1000{\text{kg/m}}^3$. If the block is in equilibrium position. The spring potential energy is $4x$. Find the value of $x$ . Round off to nearest integer.

• A. 3
• B. 3.00
• C. 3.0

Answer 1

Answer: (A), (B), (C)

Given:
Block density $\rho =2000{\text{kg/m}}^3$
mass $m=10\text{kg}$
Spring stiffness $k=100$ $\text{N/m}$.
liquid density ${\rho }_{liquid}=1000{\text{kg/m}}^3$


At equilibrium
Spring force = Weight —Buoyant force
$\therefore Kx=mg-F_B$

$Kx=mg[1-\frac{{\rho }_{liquid}}{\rho }]$

$100x=10\times 10[1-\frac{1000}{2000}]$

or, $x=0.5$ m

∴ Potential energy of spring

$\frac{1}{2}K{x}^2=\frac{1}{2}\times 100\times 0.5\times 0.5=12.5$ J

But the given Potential energy is = $4x$

$\Rightarrow 4x=12.5$

$\Rightarrow x=\frac{12.5}{4}$

$\Rightarrow x=3.125$

$\therefore x\approx 3$