Question

A cubical block of wood of edge $3cm$ floats in water. The lower surface of the cube just touches the free end of a vertical spring fixed at the bottom of the pot. Find the maximum weight that can be put on the block without wetting the new weight. Density of wood $=800kg/m^3$ and spring constant of the spring$=50N/m.$(Take $g=10m/s^2)$

Answer 1

Fraction of volume immersed before putting the new weight$=\frac{{\rho }_{block}}{{\rho }_{water}}$
$=\frac{800}{1000}$
$=0.8$
i.e., 20% of 3 cm or 0.6 cm is above water. Let W is the new weight, then spring will be compressed by 0.6 cm.
$\therefore$ W+weight of block=Upthrust on whole volume of block+spring force
or
$W={(3\times {10}^{-2})}^3\times 1000\times 10+50\times (0.6\times {10}^{-2})-{(3\times {10}^{-2})}^3\times 800\times 10$
$\therefore$ $W=0.354$ N