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AP Physics 2 : Fluids

Study concepts, example questions & explanations for AP Physics 2

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6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : Fluids

Which of the following statements best describes the buoyant force on an object submerged in water?

Possible Answers:

The buoyant force is equal to the volume of water displaced by the object.

None of these accurately describe buoyant force.

The buoyant force is equal to the weight of the water displaced by the object.

The buoyant force is equal to the weight of the object submerged in water.

The buoyant force is equal to the weight of all of the water.

Correct answer:

The buoyant force is equal to the weight of the water displaced by the object.

Explanation:

The correct answer is that the buoyant force is equal to the weight of the water displaced by the object.

It is an upward force that is exerted on the object because of the volume of fluid displaced by that object.

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Example Question #22 : Buoyant Force

$\rho_{air}=1.225\ \frac{\textup{kg}}{\textup{m}^3}$

$\rho_{Helium}=0.164\ \frac{\textup{kg}}{\textup{m}^3}$

What mass of helium would be needed to lift a $1\textup{ kg}$ balloon and a $5\textup{ kg}$ camera off of the ground?

Possible Answers:

This will be impossible

$3.082\textup{ kg}$

$0.245\textup{ kg}$

$10.35\textup{ kg}$

$0.889\textup{ kg}$

Correct answer:

$0.889\textup{ kg}$

Explanation:

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

$F_{gravity}=g*(m_{camera}+m_{balloon}+m_{Helium})$

$F_{buoyant}=V_{balloon}*\rho_{air}*-g$

$m_{Helium}=\rho_{Helium}*V_{Helium}$

$V_{Helium}=V_{balloon}$

Combining equations:

$F_{net}=g*(m_{camera}+m_{balloon}+m_{Helium})+\frac{m_{Helium}}{\rho_{Helium}}*\rho_{air}*-g$

Plugging in values:

$0=-9.8*(5+1+m_{Helium})+\frac{m_{Helium}}{.164}*1.225*9.8$

Solving for $m_{Helium}$

$m_{Helium}=0.889\textup{ kg}$

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Example Question #23 : Fluid Statics

A block weighing 150 N is suspended by a string and submerged in water. The block is times denser than water. What is the apparent weight of the block in water?

Possible Answers:

30 N

120 N

150 N

180 N

80 N

Correct answer:

120 N

Explanation:

The correct answer is 120 N

We know that the block is five times denser than the water. This means that a weight $\frac{1}{5}$ of the block would be the buoyant force which can be calculated as the following:

$\frac{150N}{5}=30N$

We can subtract the 30 N of weight from the original weight of the block which was 150 N

150 N - 30 N = 120 N

We obtain 120 N

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Example Question #21 : Buoyant Force

A block with a volume of 20 L sinks to the bottom of a water tank. What is the buoyant force on the block?

Possible Answers:

100 N

There is no buoyant force on the block

200 N

10 N

20 N

Correct answer:

200 N

Explanation:

The correct answer is 200 N 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 20 kg because 20L was displaced. The weight of 20 kg is 200 N which is equal to the buoyant force on the block.

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Example Question #21 : Fluids

An 1.2kg paint can of volume 1550ml is floating on top of a lake, but water is slowly leaking in. Determine the volume of water present in the can when it finally sinks.

Possible Answers:

.560L

.775L

.350L

None of these

1.550L

Correct answer:

.350L

Explanation:

When the can sinks, the net force of the water in the can plus gravity will be greater than the buoyant force. So the point of sinking is when they are equal.

$m_{can}*g+m_{water}*g=\rho_{water}*g*V_{can}$

Solving for $m_{water}$

$-m_{can}+\rho_{water}*V_{can}=m_{water}$

Plugging in values:

$-1.2kg+1\frac{kg}{L}*1.550L=m_{water}$

$m_{water}=.350kg$

Using the density of water,

$V_{water}=.350L$

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Example Question #26 : Fluid Statics

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

A baseball has a mass of 145g and a circumference of 230mm . Determine what percentage of a baseball will be submerged when it is floating in seawater after a home run?

Possible Answers:

$69\%$

$50\%$

$85\%$

$23\%$

None of these

Correct answer:

$69\%$

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 230mm , it is 23cm

$C=2\pi r=23cm$

r=3.66cm

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

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

$V_{ball}=205cm^3$

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

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Example Question #27 : Fluid Statics

A ship has mass 7.4*10^7kg . What volume of water would it need to be able to displace in order to be able to float?

Density of seawater: $1.03\frac{kg}{L}$

Possible Answers:

$7.18*10^{7}L$

$3.95*10^{7}L$

None of these

$4.40*10^{7}L$

$6.66*10^{7}L$

Correct answer:

$7.18*10^{7}L$

Explanation:

$F_{net}=0=F_{gravity}+F_{buoyancy}$

$0=mg+\rho_{water}*-g*V$

Solving for

$V=\frac{m}{\rho_{water}}$

$V=\frac{7.4*10^7 kg}{1.03\frac{kg}{L}}$

$V=7.18*10^{7}L$

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Example Question #28 : Fluid Statics

Determine the volume of water that needs to be displaced in order for a 90kg man to be able to float.

Possible Answers:

180L

90mL

90L

None of these

45L

Correct answer:

90L

Explanation:

When floating, the net force on the man is zero. The net force is made up of buoyant force and gravity.

$F_{net}=0=F_{gravity}+F_{buoyant}$

$0=mg-\rho*g*V$

Solving for volume

$V=\frac{m}{\rho}$

Plugging in values:

$V=\frac{90}{1}$

V=90L

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Example Question #23 : Fluid Statics

A ship has mass 4.8*10^7kg . What volume of water would it need to be able to displace in order to be able to float?

$\rho_{seawater}=1.03\frac{kg}{L}$

Possible Answers:

$4.88*10^{7}L$

$4.66*10^{7}L$

$4.11*10^{7}L$

None of these

$5.55*10^{7}L$

Correct answer:

$4.66*10^{7}L$

Explanation:

$F_{net}=0=F_{gravity}+F_{buoyancy}$

$0=mg+\rho_{water}*-g*V$

Solving for

$V=\frac{m}{\rho_{water}}$

$V=\frac{4.8*10^7 kg}{1.03\frac{kg}{L}}$

$V=4.66*10^{7}L$

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Example Question #21 : Fluids

A nitrogen bubble of radius 1cm is pumped into a bathtub from rubber tubing. Determine the net force on the bubble. The density of nitrogen is $0.0012506 \frac{grams}{cm^3}$

Possible Answers:

None of these

$9.44*10^{-6}N$

$3.66*10^{-6}N$

$4.18*10^{-6}N$

$9.12*10^{-6}N$

Correct answer:

$4.18*10^{-6}N$

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:

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

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

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AP Physics 2 : Fluids

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #21 : Fluids

Example Question #22 : Buoyant Force

Example Question #23 : Fluid Statics

Example Question #21 : Buoyant Force

Example Question #21 : Fluids

Example Question #26 : Fluid Statics

Example Question #27 : Fluid Statics

Example Question #28 : Fluid Statics

Example Question #23 : Fluid Statics

Example Question #21 : Fluids

All AP Physics 2 Resources

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