Homework 7
−
Solution
CHANG
1 Consider the energy equation for the collar’s motion up the smooth rod:
K
A
U
gA
U
eA
W
′
A
−
B
K
B
U
gB
U
eB
Note that even though the rod is frictionless there is an external force applied so
W
′
A
−
B
W
F
. The displacement is the distance traveled:
s
0.3 m, and the work done by force
F
is
W
F
F
cos
s
where
30
∘
. For the gravitational potential energy, with the datum
at the bottom of rod we have
y
A
0 and
y
B
0.3 m. The deformations in the spring at the
starting and final positions are found from geometry:
x
A
l
A
−
l
0
0.4
−
1.2
−
0.8 m
x
B
l
B
−
l
0
0.4
2
0.3
2
−
1.2
−
0.7 m
The speed can thus be calculated, with
A
0, as
1
2
m
A
2
mgy
A
1
2
kx
A
2
F
cos
s
1
2
m
B
2
mgy
B
1
2
kx
B
2
0
0
1
2
200
−
0.8
2
60
cos30
∘
0.3
1
2
5
B
2
5
9.8
0.3
1
2
200
−
0.7
2
B
2.52 m/s
CHANG
2 Take the starting point at the bottom of the hill as position 1 and the highest
point the block reaches up on the hill as position 2. There is no elastic potential energy (no
spring) so the energy equation for the block’s motion is written as
K
1
U
g
1
W
′
1
−
2
K
2
U
g
2
Work produced during the motion has two parts: work done by friction and work done by
the pushing force. Work by friction can be computed using
W
f
−
k
Ns
where
N
mg
cos
F
sin
from Newton’s 2nd law, and work by the
horizontal
force is simply
W
F
F
cos
s
where
. Using the bottom of the hill as the datum, for the gravitational
potential energy calculation we have
y
1
0 and
y
2
h
, where
h
is the maximum height the
block reaches. The displacement
s
along the ramp can be related from geometry to the height
as
y
2
h
s
sin
s
h
sin
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 Spring '08
 TSChang
 Physics, Energy, Force, Friction, Potential Energy, Work, 2 m/s, 2 m, energy equation

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