Question

A 40 cm. diameter sphere that weighs 400 n is released from rest underwater in claytor lake. its initial acceleration is about:

Answer 1

Newton's second law states that the resultant of the forces acting on the sphere is equal to the product between its mass m and its acceleration a:
$\sum F = ma$

There are two forces acting on the sphere: its weight W, directed downward, and the buoyancy B, directed upward. So Newton's second law becomes
$W-B = ma$ (1)

We already know the weight of the sphere: $W=mg=400 N$, from which we can also find the mass of the sphere:
$m= \frac{W}{g}= \frac{400 N}{9.81 m/s^2}=40.8 kg$

The buoyancy is equal to the mass of the water displaced:
$B=\rho_W V g$
where
$\rho_W = 1000 kg/m^3$ is the water density
V is the volume of water displaced, which corresponds to the volume of the sphere, since the sphere is underwater
g is the gravitational acceleration
The radius of the sphere is
$r= \frac{40 cm}{2}=20 cm=0.20 m$, so its volume is
$V= \frac{4}{3} \pi r^3 = \frac{4}{3}\pi (0.20 m)^3 = 3.55 \cdot 10^{-2} m^3$

So now we can rewrite Newton's second law (1) as
$mg-\rho_W V g = ma$
and solve it to find the acceleration of the sphere, a:
$a=g- \frac{\rho_W V g}{m}=9.81 m/s^2 - \frac{(1000 kg/m^3)(3.55 \cdot 10^{-2} m^3)(9.81 m/s^2)}{40.8 kg} =1.76 m/s^2$

and this acceleration is directed downward, since it has the same sign of g.