Practice Problems Solutions
Fall 2006  Quiz 1 on Wednesday 27 September
1a)
F
=
ma
=
m
dv
dt
=
mv
dv
dx
=
m
α
x
−
α
x
2
.
F
=
−
mα
2
/x
3
1b)
dx
dt
=
α
x
assuming
x
n
= 0,
xdx
=
αdt
Δ(
x
2
) =
α
Δ
t
limit as
x
i
→
0 gives Δ
t
=
x
2
f
/α
1c)
U
(
x
) =
−
i
Fdx
=
−
i
−
mα
2
x
3
dx
=
−
mα
2
2
x
2
+
C
Constraint
U
(
∞
) = 0 gives
C
= 0.
U
(
x
) =
−
mα
2
2
x
2
2a)
μ
s
N
≥
mg
,
N
=
ma
x
a
x
≥
g/μ
s
2b) Friction between blocks is
mg
if small block is not falling.
In ydirection, for large block :
N
−
Mg
−
mg
=
Ma
y
= 0
N
= (
M
+
m
)
g
In xdirection, for large block:
F
−
μ
k
N
−
ma
x
=
Ma
x
F
= (
M
+
m
)
a
x
+
μ
k
(
M
+
m
)
g
≥
(
M
+
m
)
g/μ
s
+
μ
k
(
M
+
m
)
g
F
≥
(
M
+
m
)
g
(
μ
k
+ 1
/μ
s
)
2c)
μ
k
N
−
mg
=
ma
y
,
N
=
ma
x
a
y
=
μ
k
a
x
−
g
3) Each “piece” of rope must be lifted to the table, requiring a work of (
dm
)
gh
where
dm
is the mass of the rope piece,
g
is the gravitational acceleration, and
h
is the height the
piece was lifted. This can be rewritten, (
dm
)
gh
= (
m
l
dh
)
gh
.
The total work is just the sum of the work for every piece.
W
=
i
d
0
mg
l
hdh
=
mg
l
1
2
h
2

d
0
=
mgd
2
/
2
l
4a)
F
=
−
kv
3
=
ma
=
m
dv
dt
=
m
dv
dx
dx
dt
=
mv
dv
dx
−
kv
2
m
=
dv
dx
−
I
k
m
dx
=
I
dv
dx
v
2
dx
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m
=
1
v
+
C
Requiring
v
=
v
0
at
x
=
x
0
= 0 allows us to fnd
C
=
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 Fall '09
 Lamb
 mechanics, Energy, Kinetic Energy, Potential Energy, dx dt dx, kxt, dv dx dx

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