Work and Kinetic Energy
63
6.7.
IDENTIFY:
All forces are constant and each block moves in a straight line. so
cos
WF
s
φ
=
. The only direction the
system can move at constant speed is for the 12.0 N block to descend and the 20.0 N block to move to the right.
SET UP:
Since the 12.0 N block moves at constant speed,
0
a
=
for it and the tension
T
in the string is
12.0 N
T
=
.
Since the 20.0 N block moves to the right at constant speed the friction force
k
f
on it is to the left and
k
12.0 N
fT
==
.
EXECUTE:
(a)
(i)
0
=
°
and
(12.0 N)(0.750 m)cos0
9.00 J
W
°
. (ii)
180
=
°
and
(12.0 N)(0.750 m)cos180
9.00 J
W
−
°
.
(b)
(i)
90
=
°
and
0
W
=
. (ii)
0
=
°
and
(12.0 N)(0.750 m)cos0
9.00 J
W
°
. (iii)
180
=
°
and
(12.0 N)(0.750 m)cos180
9.00 J
W
−
°
. (iv)
90
=
°
and
0
W
=
.
(c)
tot
0
W
=
for each block.
EVALUATE:
For each block there are two forces that do work, and for each block the two forces do work of equal
magnitude and opposite sign. When the force and displacement are in opposite directions, the work done is
negative.
6.8.
IDENTIFY:
Apply Eq.(6.5).
SET UP:
&& &&
1
⋅=⋅=
ii
jj
and
0
ij ji
EXECUTE:
The work you do is
&&
&
&
((30 N)
(40 N) ) (( 9.0 m)
(3.0 m) )
⋅=
−
⋅−
−
F
si
j
i
j
"
"
(30 N)( 9.0 m)
( 40 N)( 3.0 m)
270 N m 120 N m
150 J
−
+−
−
=
−
⋅ +
⋅ =
−
Fs
"
"
.
EVALUATE:
The
x
component of
F
"
does negative work and the
y
component of
F
"
does positive work. The total
work done by
F
"
is the sum of the work done by each of its components.
6.9.
IDENTIFY:
Apply Eq.(6.2) or (6.3).
SET UP:
The gravity force is in the
direction
y
−
, so
21
()
mg
mg y
y
−
−
"
"
EXECUTE:
(a)
(i) Tension force is always perpendicular to the displacement and does no work.
(ii) Work done by gravity is
.
mg y
y
−−
When
12
yy
=
,
0
mg
W
=
.
(b)
(i) Tension does no work. (ii) Let
l
be the length of the string.
(
)
(2 )
25.1 J
mg
Wm
g
y
ym
g
l
=−
−
EVALUATE:
In part (b) the displacement is upward and the gravity force is downward, so the gravity force does
negative work.
6.10.
IDENTIFY:
2
1
2
K
mv
=
SET UP:
65 mi/h
29.1 m/s
=
EXECUTE:
(a)
25
1
2
(750 kg)(29.1 m/s)
3.18 10 J
K
×
(b)
2
1
11
2
K
mv
=
.
2
1
22
2
K
mv
=
, with
/2
vv
=
, so
1
1
1
24
2
(/
2
)
(
)
/
4
Km
v
m
vK
=
. The change in kinetic energy is a
decrease of
3
1
4
K
.
(c)
1
2
K
K
=
.
2
constant
2
v
, so
K
K
=
.
1
2121
2
/
(65 mi/h)
/
46 mi/h
vvKK
KK
=
.
EVALUATE:
Since
2
K
v
∼
, to have half the kinetic energy the speed must be less than half of the original speed.
6.11.
IDENTIFY:
2
1
2
K
mv
=
. Since the meteor comes to rest the energy it delivers to the ground equals its original kinetic
energy.
SET UP:
4
12 km/s
1.2 10 m/s
v
×
. A 1.0 megaton bomb releases
15
4.184 10 J
×
of energy.
EXECUTE:
(a)
84
2
1
6
1
2
(1.4 10 kg)(1.2 10 m/s)
1.0 10 J
K
=×
×
=
×
.
(b)
16
15
2.4
×
=
×
. The energy is equivalent to 2.4 onemegaton bombs.
EVALUATE:
Part of the energy transferred to the ground lifts soil and rocks into the air and creates a large
crater.
6.12.
IDENTIFY:
2
1
2
K
mv
=
. Use the equations for freefall to find the speed of the weight when it reaches the ground.
SET UP:
Estimate that a person has speed 2 m/s when walking and 6 m/s when running. The mass of an electron is
31
9.11 10
kg
−
×
. In part (c) take
y
+
downward, so
2
9.80 m/s
y
a
=+
. Estimate a shoulder height of 1.6 m.
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 Spring '06
 Buchler
 Physics, Energy, Force, Kinetic Energy, Potential Energy, Work, Velocity

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