(a) a spherically shaped buoy is made from wood, which has a specific gravity of 0.6. find the acceleration of the buoy when released from rest at the bottom of a freshwater lake. (b) if the buoy starts out 10 m below the surface of the water, determine the height above the water that it will rise after it shoots out of the water.

Buoyancy = 1000 kg/m^3 * 9.81 * vol

weight = 600 kg/m^3 * 9.81 * vol

net force up = 400 * 9.81 * vol

F = m a

400 * 9.81 * vol = 600 * vol * a

a = (2/3) * 9.81
--------------------------------
work done in rise to surface
= F * 10 = 4000*9.81*vol
= Kinetic energy on exit from water
= (1/2) m v^2
= .5 * 600 * vol * v^2
so
300 v^2 = 4000* 9.81
v^2 = 130.8
v = 11.4 m/s initial speed up

Ke = Pe at stop = m g h
but oh. I did not have to do that
work done during rise = m g h directly
4000*9.81*vol = 600*vol*9.81 * h

h = (40/6) = 6 3/3 m

note g did not matter in part b

NOte also that this problem ignores the force required to accelerate the water around the sphere. This "added mass" is actually comparable to the mass of water displaced by the sphere. I bet the text writer did not even know that (hydrodymanics, not usually covered in school or college physics but very important if you are designing fishing gear)

why is work done during the rise= mgh directly?

(a) To find the acceleration of the buoy when released, we need to determine the net force acting on it. The net force is the difference between the buoy's weight and the buoyant force.

1. Weight of the buoy:
The weight of an object can be calculated using the formula:
Weight = mass * acceleration due to gravity (g)

Since we know the specific gravity of wood and the density of freshwater, we can find the mass of the buoy. The specific gravity is the ratio of the density of the object to the density of the fluid it is in.
Specific Gravity = Density of object / Density of fluid

Given:
Specific Gravity of wood = 0.6
Density of freshwater = 1000 kg/m³ (approximate value)

Density of object = Specific Gravity * Density of fluid
Density of object = 0.6 * 1000 kg/m³
Density of object = 600 kg/m³

Now, the mass of the buoy can be calculated by multiplying the density by the volume.
Mass = Density * Volume

A sphere's volume can be calculated using the formula:
Volume = (4/3) * π * r³

Since the buoy is spherically shaped, we need to know its radius to calculate the volume. The radius is not provided in the question, so we'll assume a value, for example, r = 1 meter.

Volume = (4/3) * 3.14 * (1)^3
Volume = 4.188 m³

Mass = 600 kg/m³ * 4.188 m³
Mass = 2512.8 kg (approximately)

Now that we have the mass, we can calculate the weight:
Weight = mass * g
Weight = 2512.8 kg * 9.8 m/s² (acceleration due to gravity)
Weight = 24638.24 N

2. Buoyant force:
The buoyant force is the force exerted by the fluid on the submerged object and can be calculated using the formula:
Buoyant force = density of fluid * volume of fluid displaced * g

Since the buoy is completely submerged in freshwater, the volume of fluid displaced is equal to the volume of the buoy.

Buoyant force = 1000 kg/m³ * 4.188 m³ * 9.8 m/s²
Buoyant force = 40968.24 N (approximately)

3. Net force:
Net force = Weight - Buoyant force
Net force = 24638.24 N - 40968.24 N
Net force = -16330 N

The negative sign indicates that the net force acts in the opposite direction to the weight of the buoy.

Finally, we can calculate the acceleration using Newton's second law of motion:
Net force = mass * acceleration
-16330 N = 2512.8 kg * acceleration

Solving for acceleration:
acceleration = -16330 N / 2512.8 kg
acceleration ≈ -6.5 m/s²

Therefore, the acceleration of the buoy when released at the bottom of the freshwater lake is approximately -6.5 m/s² (directed upwards).

(b) To determine the height above the water that the buoy will rise after shooting out, we can use energy considerations.

When the buoy is released from a certain depth, it has potential energy due to its position below the surface of the water. As it rises, this potential energy is converted into kinetic energy and upward motion against gravity.

Using the law of conservation of energy:
Initial potential energy = Final kinetic energy + Final potential energy

1. Initial potential energy:
The initial potential energy is given by the formula:
Initial potential energy = mass * g * h

Given:
Mass of the buoy = 2512.8 kg
Acceleration due to gravity (g) = 9.8 m/s²
Depth below the surface = 10 m

Initial potential energy = 2512.8 kg * 9.8 m/s² * 10 m
Initial potential energy = 246078.4 J (approximately)

2. Final kinetic energy:
When the buoy reaches the surface of the water, it will have converted all of its potential energy into kinetic energy.

Final kinetic energy = 0.5 * mass * v²

To calculate the final velocity (v), we can use the equation for motion in a straight line under constant acceleration:

v² = u² + 2ad

Where:
u = initial velocity (in this case, 0 since the buoy is released from rest)
a = acceleration (which we found to be approximately -6.5 m/s²)
d = distance traveled (which is the height of the water surface above the initial depth)

In this case, the height of the water surface above the initial depth is equal to the depth below the surface:
d = 10 m

v² = 0 + 2 * (-6.5 m/s²) * 10 m
v² = -130 m²/s²

Since velocity cannot be negative in this context, the negative sign indicates that our assumed direction of acceleration (upward) is incorrect. Therefore, the acceleration should be taken as a positive value.

v² = 130 m²/s²

Final kinetic energy = 0.5 * 2512.8 kg * (130 m/s)²
Final kinetic energy = 21625432 J (approximately)

3. Final potential energy:
Once the buoy shoots out of the water, it reaches its maximum height. At this point, all its kinetic energy is converted into potential energy.

Final potential energy = mass * g * h

Since there is no motion, the final kinetic energy is zero, and all the potential energy is converted into potential energy at the maximum height.

Final potential energy = 0

Using the conservation of energy equation:
Initial potential energy = Final kinetic energy + Final potential energy

246078.4 J = 21625432 J + 0 + 0

Hence, the height above the water that the buoy will rise after it shoots out is zero. The buoy will shoot out and fall back into the water without rising above it.