Version 086 – Midterm 2 – florin – (58140)
1
This
printout
should
have
25
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
A curve of radius 52
.
9 m is banked so that
a car traveling with uniform speed 47 km
/
hr
can round the curve without relying on fric
tion to keep it from slipping to its left or right.
The acceleration of gravity is 9
.
8 m
/
s
2
.
2
.
9 Mg
μ
≈
0
θ
What is
θ
?
1. 17.8004
2. 29.4517
3. 21.7496
4. 33.6484
5. 28.5481
6. 31.2492
7. 32.3251
8. 22.4695
9. 27.0
10. 18.2
Correct answer: 18
.
2
◦
.
Explanation:
Let :
m
= 2900 kg
,
v
= 47 km
/
hr
,
r
= 52
.
9 m
,
and
μ
≈
0
.
Basic Concepts:
Consider the free body
diagram for the car.
The forces acting on
the car are the normal force, the force due to
gravity, and possibly friction.
μ
N
N
N
cos
θ
m g
N
sin
θ
x
y
To keep an object moving in a circle re
quires a force directed toward the center of
the circle; the magnitude of the force is
F
c
=
m a
c
=
m
v
2
r
.
Also remember,
vector
F
=
summationdisplay
i
vector
F
i
.
Using the freebody diagram, we have
summationdisplay
i
F
x
N
sin
θ
−
μ
N
cos
θ
=
m
v
2
r
(1)
summationdisplay
i
F
y
N
cos
θ
+
μ
N
sin
θ
=
m g
(2)
(
m g
)
bardbl
=
m g
sin
θ
(3)
m a
bardbl
=
m
v
2
r
cos
θ
(4)
and
,
if
μ
= 0
,
we have
tan
θ
=
v
2
g r
(5)
Solution:
Solution in an Inertial Frame:
Watching from the Point of View of Some
one Standing on the Ground.
The car is performing circular motion with
a constant speed, so its acceleration is just
the centripetal acceleration,
a
c
=
v
2
r
.
The
net force on the car is
F
net
=
m a
c
=
m
v
2
r
The component of this force parallel to the
incline is
summationdisplay
vector
F
bardbl
=
m g
sin
θ
=
F
net
cos
θ
=
m
v
2
r
cos
θ .
In this reference frame, the car is at rest,
which means that the net force on the car
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Version 086 – Midterm 2 – florin – (58140)
2
(taking in consideration the centrifugal force)
is zero. Thus the component of the net “real”
force parallel to the incline is equal to the
component of the centrifugal force along that
direction.
Now, the magnitude of the cen
trifugal force is equal to
F
c
=
m
v
2
r
, so
F
bardbl
=
F
net
cos
θ
=
F
c
cos
θ
=
m
v
2
r
cos
θ
F
bardbl
is the component of the weight of the car
parallel to the incline. Thus
m g
sin
θ
=
F
bardbl
=
m
v
2
r
cos
θ
tan
θ
=
v
2
g r
=
(47 km
/
hr)
2
(9
.
8 m
/
s
2
) (52
.
9 m)
×
parenleftbigg
1000 m
km
parenrightbigg
2
parenleftbigg
hr
3600 s
parenrightbigg
2
= 0
.
328783
θ
= arctan(0
.
328783)
= (0
.
317649 rad)
bracketleftbigg
180 deg
π
rad
bracketrightbigg
= 18
.
2
◦
.
002
10.0 points
You do a certain amount of work on an
object initially at rest, and all the work goes
into increasing the object’s speed.
If you do
work
W
, suppose the object’s final speed is
v
.
What will be the object’s final speed if you
do twice as much work?
1.
4
v
2.
√
2
v
correct
3.
v
√
2
4.
Still
v
5.
2
v
Explanation:
W
= Δ
K
=
1
2
m v
2
∝
v
2
so
W
∝
v
2
and
v
′
2
v
2
=
2
W
W
v
′
=
√
2
v .
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Florin
 Force, Correct Answer, kg

Click to edit the document details