Physics 112 Homework #6 Solutions, Spring 2008
March 3, 2008
Problem 1
At any point in the path of the remote control car’s motion around the cylinder, it must be
accelerating toward the center at
a
radial
=
v
2
r
in order to stay in a circular path. (a) At the bottom
of the circle, the vertical forces are the weight downward and the normal force upward.
X
F
y
=
ma
y
=
N
cyl on car

W
earth on car
N
=
N
cyl on car
m
v
2
r
=
N

mg
m
v
2
r
+
mg
=
N
N
=
61
.
8
Newtons
For part (b) the normal force from the cylinder must be pointing down (still against the car)
while gravity also points down.
X
F
y
=
ma
y
=

N
cyl on car

W
earth on car

m
v
2
r
=

N

mg
m
v
2
r
=
N
+
mg
m
v
2
r

mg
=
N
N
=
30
.
4
Newtons
where N is the magnitude of the normal force; it points down (

ˆ
y
direction)
Problem 2
(a) 0: Work is given by
Work
=
´
~
F
·
~
dl
; (i) as the tension always points perpendicular to the
direction of motion,
~
T
·
~
dl
= 0
at all points, so the work done by the tension is zero for any path
along this circle.
(ii) 0: Gravity is a conservative force, so the work done by gravity depends only on the start
and end points and is equal to the negative difference in potential energies
Work
=
ˆ
~
W
·
~
dl
= (

mgh
final
)

(

mgh
initial
) =

(
U
(
h
final
)

U
(
h
initial
))
=
mg
(
h
initial

h
final
) =
mg
(0)
(b) Motion along a semicircle from the lowest point to the highest point.
0: Again, the tension does no work because it is perpendicular to the direction of motion.
25.1 J: The gravitational work done from the bottom to the top is given by
Work
=

(
U
(
h
final
)

U
(
h
initial
))
=

mg
(Δ
h
)
=

(0
.
800
kg
)
(
9
.
81
m/s
2
)
(1
.
60
m
*
2)
=

25
.
1136
J
=

25
.
1
J
1
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Problem 2’
A 4.80kg watermelon is dropped from rest from the roof of a 25.0 m tall building.
a) calculate the work done by gravity on the watermelon during its displacement from the roof
to the ground: Work = 
Work
=

Δ
U
=
U
i

U
f
=
mg
(
h
i

h
f
)
=
(4
.
80
kg
)
(
9
.
81
m/s
2
)
(25
.
0
m
)
=
1177
.
2
J
= 1180
J
b) What is the kinetic energy and the speed just before it strikes the ground?
Work
=

Δ
U
=
Δ
KE
.
Initially the watermelon was at rest, so its initial kinetic energy is zero; the final kinetic
energy is just the change, and we use
KE
=
1
2
mv
2
to find the speed.
KE
= 1180
J
=
1
2
mv
2
1180
J
=
1
2
(4
.
80
kg
)
v
2
v
=
s
1180
J
1
2
4
.
80
kg
= 22
.
18
m/s
We could also cut to the chase by starting from the potential energy; this shows that the mass
of the object does not factor into the calculations.
1
2
mv
2
=
mg
(
h
i

h
f
)
v
2
=
2
g
(
h
i

h
f
)
v
=
p
2
gh
I
=
p
2
*
9
.
81
m/s
2
*
25
m
= 22
.
15
m/s
The difference between these two numbers comes from rounding 1177.2 J to 1180 J for significant
figures in the first part.
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 Spring '07
 LECLAIR,A
 mechanics, Energy, Force, Potential Energy, Work, kg, FY

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