Sample Problems
1. A scuba student stands on the bottom of a deep swimming pool, and lifts a
bowling ball 1m above the bottom of the pool. If the bowling ball has a specific
gravity 5.0, how much time does it take for the bowling ball to hit the bottom?
(hint: find the ball’s acceleration first)
Key:
since the ball is entirely submerged. We can replace the mass of the ball by
.
Solving for a,
Now using kinematics (TNEOMS), the time of the drop becomes
=
0.5sec
using tneom 1/2at2=y= 1m
1
2. Water pressure: Find the depth to reach the following pressures under water.
Note: 1 atm = 101kPa and the units for density should by kg/m3 if using pa!
a. 1atm
b. 2 atm
c. 3 atm
key:
KPa = 1000 Newtons/m2
Pbelow = 1 atm + gh where = 1000 kg/m3 is the density of water
P
= 101,000 + 1000 kg/m3 * 10 *h
a. h = 0
b. P = 2 atm = 202000 Pa = 101,000 + 1000 *10*h
h=10.1 m
c. P = 3 atm = 303,000Pa = 101,000Pa + 1000 *10*h
h=20.2 m
d. P = 10 atm = 1010,000 Pa = 101000 pa + 1000 *10*h
h=90.9 m
d. 10 atm (why not? As long as you don’t come up too fast & get the bends!)
2
3. a toy boat has an empty mass of 210 g and a volume of 280 cm3.
D
How much water must you add to make it sink?
Set the density of the sub equal to density of water, 1 g/cm3.
To make this submarine sink, its mass, equal to
, should be equal
3
3
to the mass of water displaced: (1 g/cm )(280 cm ) = 280 g.
Its empty mass is 210 g, so add 70 g (= 70 cm3, or 70 mL) of water.
4. Prove only about 10% of a floating ice berg is visible.
The density of ice is 0.9167 g/cm³ at 0°C, water has a density of 0.9998 g/cm³
Weight of ice = weight of water displaced
Mg = mg and
m= density*volume
Mg= Density of ice* total volume of ice = .9167*V
mg=density of water *volume of water displace
= .9998 * % underwater * V
.9167/.998*V = % underwater*V = 90%
so 90% underwater, 10% above! (Penguins rule)
3
5. a 200 g mass has a volume of 50 cm3, and hangs from
a spring. What is the effective weight (that is what will
the spring scale read) when the mass (none of the
spring) is completely submerged in water?
The mass is in equilibrium, so
. ( is the tension in
the rope, which is also the force probe reading.)
now substitute for the buoyant force
4
6. A hydraulic lift consists of a cylindrical piston of radius 2cm, which when pushed upon
with a force of only 100N is able to lift a 2000kg car sitting on a much larger cylindrical
piston. What is the area of the larger piston ?
Pressure felt same everywhere if at same height & speed
P = F/A = F/A
Fbig /F = Abig /A
Flarge/Fsmall = wt of car/ pushing force = mg/100 = 2000*10 /100 = 200
Alarge/Asmall = Area/ *(2)2 = 200 so A = .25 m2 or r = 28.3cm
7. Blood flows from your heart (.009m radius for the aorta) at about 0.33m/s. It then
travels through your capillary system at a much smaller speed of 0.00034 m/s. What is
the effective cross-sectional area of your capillary system? (assume no blood loss!)
Av = Av since mass is not lost and density is not changed
Aorta * v = *r2 *v = *(.009)2(0.33)
Acap *v = A* v
=*(3.4e -4)2so A = 0.25m2
5
8. How fast will the water come out of a 2 liter bottle with a hole 5m from the ground and
the water filled to 25cm? (assume top of water and side of water at atmospheric pressure)
Water level
hole
Pascal principle: Pressure at top same as at bottom since exposed to air
P1 = P2 = 1 atm or 105 Pa
Initial velocity at top is at rest so v1 = 0
Bournoulli:
P1
+ gh1 + ½  v12 = P2 + gh2 + ½  v22
+ gh1 + ½  (0)2 = P1 + g()) + ½  v22
gh1 = ½  v22
v2 = square root (2gh1) =√2*10*0.20m = 2m/s
P1
6
AP Problems
50m to top of well
35m deep well
9. A pump, submerged at the bottom of a well that is 35 m deep, is used to pump water
uphill to a house that is 50 m above the top of the well. The density of water is 1,000
kg/m3. All pressures are gauge pressures. Neglect friction, turbulence, and viscosity.
a. Residents of the house use 0.35 m3 of water per day. The day's pumping is completed in 2 hours during the day.
i. Calculate the minimum work required to pump the water used per day (300,000 J)
Work = PE = total wt * height= 1000kg/m3 * .35m3 * 10m/s2 * (50 + 35 m) = 300,000 J
ii.
b.
Calculate the minimum power rating of the pump. (41 Watts)
P = work/time (sec) = 300,000 J/ (2hrs * 3600 sec) = 41 watts
The average pressure the pump actually produces is 9.20 x 10 5 N/m2. Within the well the water flows at 0.50
m/s and the pipe has a diameter of 3.0 cm. At the house the pipe diameter is 1.25 cm.
i.
Calculate the flow velocity when a faucet in the house is open. (2.88 m/s)
radius = diameter /2 used in area of circle =  (radius)
Area * velocity at house = area * velocity at pump
*(.0125m/2) 2 * v = *(.03m/2)2 * 0.5m/s
v =2.88 m/s
ii.
Explain how you would calculate the minimum pressure at the faucet.
+ gh1 + ½  v12 = P2 + gh2 + ½  v22
9.2x105 + 1000*10*85 + ½*1000* (0.5)2 = P + 0 + ½ 1000 (2.88)2
solve for P since given all info (height = 0 at top of house for fluids)
P1
7
10.
10.
a.
An unoccupied, 10 kg wooden raft made of driftwood with specific gravity 0.60 floats on the surface
3
of a calm pond. [The density of water is 1,000 kg/m .]
3
What is the volume of the raft? (0.0017 m )
3
density=mass/volume so v= mass/density = 10kg/ (0.6*1000 kg/m ) = .017m
b.
3
Determine the buoyant force acting on the raft while it floats. (100 N)
2
Since floats, no net force so buoyant force = wt = mg = 10kg *10m/sec = 100N
3
Now a student under the water hangs brass (density = 8,000 kg/m ) masses on a light string attached to the raft, as
shown in the diagram above. The student adds brass a little at a time until the moment when the raft becomes
completely submerged.
c.
Determine the new buoyant force on the raft. (170N)
buoyant force = wt water displaced = density water * volume of raft and hangers (negligible) * g
3
3
2
= 1000 kg/m * .017m * 10m/sec = 170N
d.
Calculate the minimum mass of brass that the student must hang from the raft to allow the raft to become
completely submerged. (70 plug 8 = 78kg total)
If neglect volume of brass adding to the buoyant force, then
total weight (100N raft + brass) ≥ buoyant force = (170N )
so brass can weigh= 70N = m*10 and mass of brass = 7kg (but this is an underestimate)
extra buoyant force = water wt displaced by brass = volume of brass*water density *10m/s 2
100 + wt. brass = 170N + extra buoyant force
3
Vbrass*8000kg/m *10 = 70 + Vbrass * 1000kg/m3 *10
Vbrass (7000 *10) = 70 so Vbrass =.001m3
3
3
and mass of brass = density * volume = 8000 kg/m * .001m = 8kg
e.
If instead the 10 kg raft were made of freshly cut wood with specific gravity 0.80, how would the answer to
part (d) change? Check one box and justify your answer briefly.
X Less brass would be necessary since denser wood takes less space, there’s less buoyant force to hold up the
same weight of wood (same mass, so same weight). Need less extra weight to sink it !
8
11.
Derive an equation for the velocity of the stream using the variables in the picture:
energy: exit hole: KE = 1/2mV11, PE = mgy1
top :
KE= 1/2mV22, PE = mgy2
Set equal: 1/2mV22 + mgy2 = 1/2mV12 + mgy1
1/2mV12 = 1/2mV22 + mg(y2-y1)
V12 = V22 + 2g(y2-y1)
12.
What is the lifting force of an airplane wing having an area of 24m2 if the air flows 251
m/s above the wing, and 225 m/s below the
wing? Assume dens)ity of air is 1.29kg/m3.
F = pressure * area
Fnet = pressure at bottom – top * area
pressure + 1/2pv2 = pressure + 1/2pv2
(Bernoulli with h being same about)
change in pressure =
1/2p(v2 – v2) = ½ * 1.29 * (2252-2512)
Fnet = P*A = answer*24= 192,000N
9
13.
A venture pump is basically a pipe that starts wide and ends narrow. The pressure is
lower at the narrow end and can be used to pull a vacuum to “suck” out air or liquids.
Suppose the wider end has an area of .07 m2 and a pressure of 200 Pa, and the narrow
end (middle section of picture) has an area of .05 m2 and a pressure of 80 Pa. The fluid
has a density of 1.3 kg/m3. .
a. What is the pressure difference between the two ends? (120 Pa)
P= 200-80 = 120Pa
b. What is the ratio of speeds between the two ends, ie. what is V1/V2? (1.4-wrong!)
mass/sec = constant = density * area * speed (faster in shorter area)
area * velocity = area * velocity
pi*(.07)2 Velocity = pi (.05)2* velocity
(V1/V2 ) proportional to A2/A1 = (.05) / (.07) = .71
c. Use Bernoulli’s equation to solve for V2. (14m/s is V1, V2 = 19.4)
don’t know either speed so need 2 equations!
P + pgh + 1/2pv2 = P + pgh + 1/2pv2 and use V1 = V2 *A2/A1 = V2*.71
200 + ½*1.3*(V1)2 = 80 + ½*1.3*(V2)2
200 + ½*1.3*(.71*V2)2 = 80 + ½*1.3*(V2)2
(sub in for V1)
120= ½* 1.3 * (V2)2 ( 1- .712) ……………solve and get V2 = 19.4 m/s
d. What is volume flow rate Q1 in m3/sec at the initial larger end? (0.98m3/sec-wrong)
Q = mass/sec = pAv1……oops they want volume not mass! so no density, just AV1
v1 = 0.71 x V2 = 0.71* 19.4 = 13.7………………still ok
then Q = 1.3 * (.07)2*13.7 = .27……oops, no use Av=(.07)2*13.7=.21
e. What is the volume flow rate Q2 in m3/sec at the narrow section? (trick question!)
same!
10
11