HW Solutions # 7 - 8.01 MIT - Prof. Kowalski
Potential Energy Curves, Momentum and Center of Mass.
1
)
7.61
We use conservation of energy
K
i
+
U
i
+
W
other
=
E
i
+
W
other
=
E
f
=
K
f
+
U
f
(1)
1
2
mv
2
1
+
mgy
1
+
W
other
=
1
2
mv
2
2
+
mgy
2
(2)
I will measure the gravitation potential energy with respect to the
ground.
a
)
First obtain the kinetic energy - denote it as
K
h
- as if he stepped
off the platform: When he stepped of a distance h
≤
d,
W
other
=0
because of negligible air resistance the energy conservation implies:
0 +
mgh
=
K
h
+ 0
(3)
For the case he slides down the pole
W
other
after moving down the
distance d is just
f
ave
d
cos 180
o
(force opposes the motion
⇒
W is
<
0):
W
other
=
−
f
ave
d
(4)
The next step is to calculate Kinetic energy when he slides down
the pole and as it’s said make it equal to
K
h
:
0 +
mgd
+
W
other
=
K
d
+ 0
(5)
Combining equation (4) with equations (2) and (3) and
K
d
=
K
h
:
mgd
−
f
ave
d
=
mgh
(6)
Therefore:
f
ave
=
mg
(1
−
h
d
)
1

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When
h
=
d, f
= 0 ; as expected: when there is no friction sliding
down the pole has the same effect as stepping off the platform.
And when h=0 we expect
f
ave
=
mg
which is consistent with what
we derived.
b
)
Plugging the numbers give in the problem into the boxed equa-
tion:
f
ave
= 75
.
80
×
(1
−
1
2
.
5
) = 441 N
.
(7)
c
)
Let’s denote
v
at position
y
as
v
(
y
) Again setting up the equation
(1) -energy conservation- :
v
1
= 0
y
1
=
d
W
other
=
−
f
ave
(
d
−
y
)
(8)
v
2
=
v
(
y
)
y
2
=
y
(9)
∴
0 +
mgd
−
mgf
ave
(
d
−
y
) =
1
2
mv
(
y
)
2
+
mgy
(10)
1
2
mv
(
y
)
2
= (
mg
−
f
ave
)(
d
−
y
)
(11)
Using the boxed equation for
f
ave
we get:
1
2
mv
(
y
)
2
=
mg
(
h
d
)(
d
−
y
) =
mgh
(1
−
y
d
)
(12)
From which
v
(
y
) =
2
gh
(1
−
y
d
)
When
y
= 0
, v
=
√
2
gh
, which is the original condition.
When
y
=
d, v
= 0: the firman is at the top of the pole.

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