Answer
The initial acceleration of the piece of metal is $7.84~m/s^2$ downward
Work Step by Step
Since the specific gravity of the metal is 5.0, the density is $\rho = 5.0~g/cm^3$. We can express the density of water $\rho_w$ in terms of the density of the metal $\rho$:
$\frac{\rho_w}{\rho} = \frac{1.0~g/cm^3}{5.0~g/cm^3}$
$\rho_w = 0.20~\rho$
According to Archimedes' principle, the buoyant force is equal to the weight of the water that is displaced. Let $V$ be the volume of the metal. We can find the initial acceleration of the metal:
$\sum F = ma$
$mg-F_b = ma$
$\rho~V~g-\rho_w~V~g = ma$
$\rho~V~g-0.20~\rho~V~g = ma$
$0.80~\rho~V~g = ma$
$0.80~mg = ma$
$a = 0.80~g$
$a = (0.80)(9.80~m/s^2)$
$a = 7.84~m/s^2$ downward
The initial acceleration of the piece of metal is $7.84~m/s^2$ downward
