Buoyant Force

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AP Physics 2 › Buoyant Force

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Seawater density: $1.03\frac{g}{cm^3}$

A baseball has a mass of and a circumference of . Determine what percentage of a baseball will be submerged when it is floating in seawater after a home run?

CORRECT

None of these

Explanation

$F_{net}=F_{gravity}+F_{buoyant}$

$0=\rho_{seawater}*V_{submerged}*g-m*g$

Solving for $V_{submerged}$

$\frac{m_{ball}}{\rho_{seawater}}=V_{submerged}$

Plugging in values

$\frac{145}{1.03}=V_{submerged}$

$V_{submerged}=141cm^3$

If the circumference is , it is

$C=2\pi r=23cm$

$V_{ball}=\frac{4}{3}\pi r^3$

$V_{ball}=\frac{4}{3}\pi *3.66^3$

$V_{ball}=205cm^3$

$\frac{V_{submerged}}{V_{ball}}=\frac{141cm^3}{205cm^3}=69%$

What is the net force on a ball of mass and volume of when it is submerged under water?

$g = 10\frac{m}{s^2}$

$\rho_{water} = 1.0\frac{g}{ml}$

CORRECT

Explanation

The buoyant force on the ball is simply the weight of water displaced by the ball:

$F_b = V_{ball}\cdot\rho_{water}\cdot g = (0.2m^3)(1000\frac{kg}{m^3})(10\frac{m}{s^2}) = 2000 N$

The force of gravity on the ball is:

$F_g = mg = 15kg (10\frac{m}{s^2})= 150 N$

These forces oppose each other, so we can say:

$F_{net} = F_b - F_g = 2000N - 150N = 1850N$

A block of mass and volume is held in place under water. What is the instantaneous acceleration of the block, and in what direction, when it is released?

$g = 10\frac{m}{s^2}$

$\rho_{water}=1.0\frac{kg}{L}$

$3\frac{m}{s^2}, down$

CORRECT

$3\frac{m}{s^2},up$

$6\frac{m}{s^2},down$

$6\frac{m}{s^2},up$

$0\frac{m}{s^2}$

Explanation

According to Archimedes's principle, when the block is placed under water, 3.5L of water are displaced. We can then calculate the buoyant force provided by the water:

$F_{buoyant} = m_{displaced}g$

Where:

$m= \rho_w V = \left ( 1.0\frac{kg}{L}\right )(3.5L) = 3.5kg$

Plugging in our values to the first expression:

$F_b = (3.5kg)\left ( 10\frac{m}{s^2}\right )$

Then we can use Newton's 2nd law to determine the acceleration of the block:

$F_{net}=ma$

There are two forces acting on the block, gravity and buoyancy, and they are in opposite directions. If we designate a downward force as positive, we get:

Substituting in the expression for force due to gravity:

Rearranging for acceleration:

$a = \frac{mg-F_b}{m}$

Substituting in values:

$a = \frac{ (5kg)\left ( 10\frac{m}{s^2}\right )-35N}{5kg}$

$a = 3\frac{m}{s^2}$

A balloon of mass $45\textup{ g}$ is inflated to a volume of $2\textup{ L}$ with pure . Determine the buoyant force it will experience when submerged in water.

$19.6\textup{ N}$

CORRECT

None of these

$15\textup{ N}$

$28.1\textup{ N}$

$3.4\textup{ N}$

Explanation

Use the equation for buoyant force:

$F_b=\rho*|g|*V$

Where

$\rho$ is the density of the medium

is the acceleration due to gravity

is the volume

Plugging in values:

$F_b=\frac{1kg}{L}*|-9.8\frac{m}{s^2}|*2L$

A nitrogen bubble of radius is pumped into a bathtub from rubber tubing. Determine the net force on the bubble. The density of nitrogen is $0.0012506 \frac{grams}{cm^3}$

$4.18*10^{-6}N$

CORRECT

None of these

$3.66*10^{-6}N$

$9.44*10^{-6}N$

$9.12*10^{-6}N$

Explanation

$F_{net}=F_{gravity}+F_{buoyant}$

$F=mg-\rho *g*V$

The mass of the bubble will also be related to density and volume:

$F=V\rho_{nitrogen}g-\rho_{water} *g*V_{bubble}$

Plugging in values:

$F=(\frac{4\pi}{3}*.01^3)*.0012506*-9.8-(\frac{4\pi}{3}*.01^3)*1*-9.8$

$F=4.18*10^{-6}N$

A semi-hollow, spherical ball with an empty volume of is submerged in water and has an initial mass of . The ball develops a leak and water begins entering the ball at a rate of $0.6\frac{L}{min}$ . How long does it take before the buoyant force on the ball is equal to the gravitational force?

$g = 10\frac{m}{s^2}$

$\rho_w = 1000\frac{kg}{m^3}$

CORRECT

Explanation

We are asked when:

$m_bg = m_{w,displaced}g$

$m_b = m_{w,displaced}$

$m_b = \rho_wV_b$

Now we need to develop an expression for the mass in the ball using the rate at which water enters the ball:

$m_b = m_i + \dot{m}t$

Where:

$\dot{m} = \dot{V}\rho_wt$

$m_b = m_i + \dot{V}\rho_wt$

Plugging this into expression (1):

$m_i + \dot{V}\rho_wt= \rho_wV_b$

Rearranging for time, we get:

$t= \frac{\left ( \rho_wV_b- m_i \right )}{\dot{V}\rho_w}$

Plugging in our values, we get:

$t= \frac{\left ( 1\frac{kg}{L}\right )(0.8L)- (0.4kg)}{\left ( 0.6\frac{L}{min}\right )\left ( 1\frac{kg}{L}\right )}$

$t = \frac{2}{3}min = 40s$

A block of mass is sinking in water at a constant velocity. There is a constant drag force of acting on the block. What is the volume of the block?

$g = 10\frac{m}{s^2}$

$\rho_w = 1000\frac{kg}{m^3}$

CORRECT

Explanation

We will start with Newton's 2nd law for this problem:

$F_{net}=ma$

Since the block is traveling at a constant velocity we can say:

$F_{net}=0$

There are 3 forces acting on the block: gravitational, buoyant, and drag force. If we denote a downward force being positive, the expression becomes:

Where:

$F_g = m_{block}g$

$F_b = m_{w,displaced}g$

Substituting these in:

$m_{block}g - m_{w,displaced}g - 3N = 0$

Where according to Archimedes's principle:

$m_w = \rho_wV_b$

Plugging this in:

$m_{block}g - \rho_w V_bg - 3N = 0$

Rearranging for volume:

$m_{block}g - 3N = \rho_w V_bg$

$V_b = \frac{m_{block}g - 3N }{\rho_w g}$

Plugging in values:

$V_b =\left (\frac{(2kg)\left ( 10\frac{m}{s^2}\right )-3N}{\left ( 1000\frac{kg}{m^3}\right )\left ( 10\frac{m}{s^2}\right )} \right )\left ( \frac{1000L}{1m^3}\right )$

A block with a volume of sinks to the bottom of a water tank. What is the buoyant force on the block?

CORRECT

There is no buoyant force on the block

Explanation

The correct answer is because although the block has sunk, there is still a buoyant force. This buoyant force is the result of the block displacing a volume of water, equal to the block's volume. The weight of the water volume displaced is because was displaced. The weight of is which is equal to the buoyant force on the block.