Question

A ball of mass $10\text{kg}$ and density $1{\text{gm/cm}}^3$ is attached to the base of a container having a liquid of density $1.1{\text{gm/cm}}^3$, with the help of a spring of spring constant $200\text{N/m}$ as shown in the figure. If the container starts going up with an acceleration $2{\text{m/s}}^2$ , the elongation in the spring (in cm, upto two decimals) is -

Answer 1

Equating forces at equilibrium
$F_b-kx-m(g+a)=0$
$kx=F_b-m(g+a)$
$kx=\left(\frac{10}{{10}^3}\right)(1.1\times {10}^3)\times (10+2)-10\times (10+2)$
$=120\times 0.1$
$x=\frac{12}{200}m=6\text{cm}$