Question

A 1.14 kg hollow ball with a radius of 0.133 m, filled with air, is released from rest at depth of 2.09 m in a pool of water. (depth is to the center of ball) What is the net vertical force acting on the ball?(Neglect all frictional effects. Neglect the ball's motion when it is only partially submerged. Neglect the mass of the air in the ball.) What is the work done by the net vertical force on the ball as the ball moves from the bottom of the pool to the surface? How high above the water does the ball shoot upward?

Answer 1

The net force is the difference between the upward buoyant force and the downward weight of the ball. The buoyant force is the product of the density (ρ) of the fluid, in this case water, the volume (V) of the ball, and the acceleration of gravity. The weight of the ball is its mass times gravity, so:

∑F = B - w
= ρ(water)V(ball)g - mg
= (1000kg/m³)[4π(0.133m)³/3)(9.80m/s²)] - (1.16kg)(9.80m/s²)
= 30.74N - 11.36N = 19.38N

Since work is force applied over distance:

W = FΔy
= (19.38N)(2.09m)
= 40.502J

To find the height the ball gets to above the surface, you first need the speed of the ball at the point where it emerges from the surface, but you'll also need the acceleration of the ball. From Newton's 2nd law:

F = ma
a = F / m
= 19.38N / 1.14kg
= 17m/s²

The velocity at the water's surface is v in this kinematics equation:

v² = v₀² + 2aΔy
= 0 + 2(17m/s²)(2.09m)---->take the square root of this
v = 8.42m/s

When the ball emerges it has speed 8.42m/s and is under the acceleration of gravity, the acceleration under the water's surface no longer applies. Using the same equation (where v above is now v₀ and v is zero because it comes to a stop at the top of its flight) and solving it for Δy:

Δy = v₀² / 2g
= (8.42m/s)² / (2 x 9.80m/s²)
= 3.617m

Hope this helps :)

Calculations maybe faulty, but logic is correct.