A block of density 2000 kg/m3 and mass 10 kg is suspended by a spring stiffness 100 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 kg/m3. If the block is in equilibrium position then,
the magnitude of buoyant force acting on the block is 50N.
the spring potential energy is 12.5 J.
At equilibrium
kx=Mg=FB
kx=mg(1−σρ)
⇒ 100(x)=100(1−10002000)
⇒ x=0.5 m
FB=50 N
P.E.=12kx2=12.5 J