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Work
111
11.6.
Solve:
(a)
x
Éº
ª*
( 5.0
4.0 ) N (3.0 ) m
( 15.0
12.0
) J= 15.0 J.
W
F
r
i
j
i
i i
j i
=
⋅∆ = 
+
⋅
= 
⋅ +
⋅

r
r
r
(b)
x
É
≡♠
( 5.0
4.0 ) N ( 3.0 ) m
(15.0
12.0
) J= 12.0 J.
W
F
r
i
j
j
i j
j j
=
⋅∆ = 
+
⋅ 
=
⋅ 
⋅

r
r
r
11.13.
Model:
Use the workkinetic energy theorem to find velocities.
Visualize:
Please refer to Figure Ex11.13.
Solve:
The workkinetic energy theorem is
f
i
f
x
2
2
f
i
i
f
x
2
2
2
f
f
0 m
1
1
area under the force curve from
to
2
2
1
1
1
(0.500 kg)(2.0m/s)
1.0 J
area from 0 to
2
2
2
x
x
x
K
mv
mv
W
F dx
x
x
mv
mv
F dx
x
∆
=

=
=
=
⇒

=

=
=
∫
∫
2
f
f
2
f
f
2
f
f
1
At =1 m:
(0.500 kg)
1.0 J
12.5 J
7.35 m/s
2
1
At
2 m:
(0.500 kg)
1.0 J
20 J
9.17 m/s
2
1
At
3 m:
(0.500 kg)
1.0 J
22.5 J
9.70 m/s
2
x
v
v
x
v
v
x
v
v

=
⇒
=
=

=
⇒
=
=

=
⇒
=
11.40.
Model:
Model the rock as a particle, and apply the workkinetic energy theorem.
Visualize:
Solve:
(a)
The work done by Bob on the rock is
2
2
2
2
Bob
1
0
1
1
1
1
1
(0.50 kg)(30 m/s)
225 J
2
2
2
2
W
K
mv
mv
mv
= ∆
=

=
=
=
(b)
For a constant force,
Bob
Bob
Bob
Bob
/
225 N
W
F
x
F
W
x
=
∆ ⇒
=
∆ =
.
(c)
Bob’s power output is
Bob
Bob rock
P
F v
=
and will be a maximum when the rock has maximum speed. This is just
as he releases the rock with
v
rock
=
v
1
=
30 m/s. Thus,
max
Bob 1
6750 W
6.75 kW.
P
F v
=
=
=
11.48.
Model:
Model the two blocks as particles. The two blocks make our system.
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112
Visualize:
We place the origin of our coordinate system at the location of the 3.0 kg block.
Solve:
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 Spring '08
 Nandi
 Physics, Energy, Kinetic Energy, Work

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