Midterm 2
Physics 1200 – Humanic
XX:XX pm lecture
March 29, 2017
Name _________________________________Rec. Instructor_____________________
* Exam is closed book and closed notes. A calculator is allowed.
* Write your name and the name of your recitation instructor on every page of the exam.
* This exam consists of two sections:
Section I – 8 multiple choice questions (40 points)
Section II – Two problems where you are required to show work (50 points)
* You have 55 minutes to complete both sections.
PEgravity = mgh PEspring = 12 kx 2
!
!
Pf = P0
! ! !
P = p1 + p 2
KE = 12 mv 2 WNC = E f − E0
!
! !
p = mv J =
!
(∑F) Δt = p!
f
!
− p0
s = rθ ω =
∑τ
EXT
Δθ
Δt
α=
= Iα τ = Fl
I SOLID SPHERE = MR
2
5
2
Δω
Δt
v01 − v02 = − ( v f 1 − v f 2 )
vT = rω
I = ∑ ( mr 2 ) I DISK = 12 MR 2
WR = τθ
KER = Iω
!
xCM =
m1 x1 + m2 x2
m1 + m2
2
P=
W
t
E f = E0
g = 9.80 m/s2 W = (F cosθ )s
vCM =
m1v1 + m2 v2
m1 + m2
m1v 201 + 12 m2 v 202 = 12 m1v 2f 1 + 12 m2 v 2f 2
aC = rω 2 1 rev = 2π rad = 360 o
aT = rα
1
2
1
2
!
∑ F = ma
ω = ω 0 + α t θ = 12 (ω 0 + ω )t ω 2 = ω 02 + 2αθ θ = ω 0 t + 12 α t 2
m1v01 + m2 v02 = m1v f 1 + m2 v f 2
E = KE + PE


L = Iω
WR = ΔKER


L f = L0
I THIN RING = MR 2
(∑τ ) Δt = ΔL
EXT
E = Iω + Mv 2 + Mgh E f = E0
1
2
2
1
2
Chapter 10
Fluids
Mass Density
DEFINITION OF MASS DENSITY
The mass density of a substance is the mass of a
substance divided by its volume:
m
ρ=
V
SI Unit of Mass Density: kg/m3
Mass Density
Mass Density
Example: Blood as a Fraction of Body Weight
The body of a man whose weight is 690 N contains about
5.2x10-3 m3 of blood.
(a) Find the blood’s weight and (b) express it as a
percentage of the body weight.
m = ρV =(1060 kg m 3 ) ( 5.2 ×10 −3 m 3 ) = 5.5 kg
Mass Density
(
2
)
(a)
W = mg = (5.5 kg ) 9.80 m s = 54 N
(b)
54 N
Percentage =
×100% = 7.8%
690 N
Pressure
Definition of Pressure: The pressure, P, exerted by
a fluid is defined as the magnitude of the force, F,
acting perpendicular to a surface divided by the area,
A, over which the force acts.
F
P=
A
SI Unit of Pressure: 1 N/m2 = 1Pa
Pascal
The pressure of the moving air molecules
(blue dots) acting inside of a tire.
Pressure
Example: The Force on a Swimmer
Suppose the pressure acting on the back
of a swimmer’s hand is 1.2x105 Pa. The
surface area of the back of the hand is
8.4x10-3m2.
(a) Determine the magnitude of the force
that acts on it.
(b) Discuss the direction of the force.
Pressure
P=
F
A
(
)(
F = PA = 1.2 ×10 5 N m 2 8.4 ×10 −3 m 2
= 1.0 ×10 3 N
Since the water pushes perpendicularly
against the back of the hand, the force
is directed downward in the drawing.
)
Pressure
Atmospheric Pressure at Sea Level: 1.013x105 Pa = 1 atmosphere
Pressure and Depth in a Static Fluid
Consider a static fluid, i.e. a fluid at
rest:
!
a=0 ⇒
∑F
y
!
∑F = 0
= P2 A − P1 A − mg = 0
P2 A = P1 A + mg
m = ρV
Pressure and Depth in a Static Fluid
V = Ah
P2 A = P1 A + ρ Vg
P2 A = P1 A + ρ Ahg
P2 = P1 + ρ gh
Pressure and Depth in a Static Fluid
Conceptual Example: The Hoover Dam
Lake Mead is the largest wholly artificial
reservoir in the United States. The water
in the reservoir backs up behind the dam
for a considerable distance (120 miles).
Suppose that all the water in Lake Mead
were removed except a relatively narrow
vertical column.
Would the Hoover Dam still be needed
to contain the water, or could a much less
massive structure do the job?
Pressure and Depth in a Static Fluid
Example: The Swimming Hole
Points A and B are located a distance of 5.50 m beneath the surface
of the water. Find the pressure at each of these two locations.
Pressure and Depth in a Static Fluid
P2 = P1 + ρ gh
atmospheric pressure
$
!
!#!!
"
P2 = 1.01 ×10 5 Pa + 1.00 ×10 3 kg m 3 9.80 m s 2 (5.50 m )
(
= 1.55 ×10 5 Pa
) (
)(
)
Pascal’s Principle
PASCAL’S PRINCIPLE
Any change in the pressure applied
to a completely enclosed fluid is transmitted
undiminished to all parts of the fluid and
enclosing walls.
A2
Pascal’s Principle
Situation in the Figure – The plunger and
piston are at the same height.
P2 = P1 + ρ g (0 m)
F2 F1
=
A2 A1
⎛ A2 ⎞
F2 = F1 ⎜⎜ ⎟⎟
⎝ A1 ⎠
A2
Pascal’s Principle
Example: A Car Lift
The input piston has a radius of 0.0120 m
and the output plunger has a radius of
0.150 m.
The combined weight of the car and the
plunger is 20500 N. Suppose that the input
piston has a negligible weight and the bottom
surfaces of the piston and plunger are at
the same level.
What is the required input force of the piston
In order to maintain the position of the car
and plunger?
A2
Pascal’s Principle
! A2 $
F2 = F1 # &
" A1 %
" A1 %
⇒ F1 = F2 $ '
# A2 &
A2
F1 = ( 20500 N )
π ( 0.0120 m )
π ( 0.150 m )
2
2
= 131 N