Question

A light unstretched/uncompressed spring with spring constant k rests vertically on the bottom of a beaker that contains a fluid of density ?_f ?(Figure 1) . A block of wood with mass m and density ? is connected to the spring and the spring-block system is allowed to come to static equilibrium. (?_f> ?)

A) Write an expression for the elongation distance d of the spring in the new equilibrium position. Write your answer in terms of m, k, ?_f, ?, and any aprropriate constants.

Answer 1

forces acting on block of mass m

weight of the body Fw=mg acting down wards

the volume of block v, we know that $volume=\frac{mass}{density}$

   $v=\frac{m}{\rho }$

the buoyancy force acting on the block $Fb=\rho _{f}gv$ $=\rho _{f}g\frac{m}{\rho }$

$=\frac{\rho _{f}mg}{\rho }$ acting upwards

if the elongation of the spring x=d and spring constant k

the spring force Fs=kx=kd develop opposite to elongation (down wards)

the given system is in equilibrium.then $Fb=Fs+Fw$

$\frac{\rho _{f}mg}{\rho }=kd+mg$

$kd=\frac{\rho _{f}mg}{\rho }-mg$

$kd=mg\left ( \frac{\rho _{f}}{\rho }-1 \right )$

the elongation of spring $d=\frac{mg}{k}\left ( \frac{\rho _{f}}{\rho }-1 \right )$