sousse (res2468) – HW 5 – Kleinman – (58225)
1
This
printout
should
have
28
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
This HW covers Chs.
7 and 8. Note that
multiple choice questions with fewer than 8
answer choices have been reduced in value to
only 5 points.
001
10.0 points
A woman weight lifter can lift a 130 lb weight
from the floor to a stand 3
.
2 ft off the ground.
What
is
the
total
work
done
by
the
woman?
Correct answer: 564
.
096 J.
Explanation:
Let :
F
= 130 lb
and
d
= 3
.
2 ft
.
W
=
F d
= (130 lb) (3
.
2 ft)
×
1
.
356 J
ft
·
lb
=
564
.
096 J
.
002
10.0 points
A catcher “gives” with a baseball when catch
ing it.
If the baseball exerts a force of 464 N on
the glove such that the glove is displaced 13.2
cm, how much work is done by the ball?
Correct answer: 61
.
248 J.
Explanation:
Basic Concept:
W
net
=
F
net
d
cos
θ
=
F
net
d
since
θ
= 0
◦
⇒
cos
θ
= 1.
Given:
F
net
= 464 N
d
= 13
.
2 cm
Solution:
W
net
=
F
net
d
= (464 N)(0
.
132 m)
= 61
.
248 J
003
(part 1 of 2) 10.0 points
A 2 kg object is given a displacement Δ
vectors
=
(3 m)ˆ
ı
+ (1 m)ˆ
+ (
−
4 m)
ˆ
k
along a straight
line.
During the displacement, a constant
force
vector
F
= (4 N)ˆ
ı
+ (
−
5 N)ˆ
+ (5 N)
ˆ
k
acts on
the object.
Find the work done by
vector
F
for this displace
ment.
Correct answer:
−
13 J.
Explanation:
Let :
m
= 2 kg
,
Δ
vectors
= (3 m)ˆ
ı
+ (1 m)ˆ
+ (
−
4 m)
ˆ
k ,
and
vector
F
= (4 N)ˆ
ı
+ (
−
5 N)ˆ
+ (5 N)
ˆ
k .
The work is
W
=
vector
F
·
Δ
vectors
=
F
x
s
x
+
F
y
s
y
+
F
y
s
z
= (4 N) (3 m) + (
−
5 N) (1 m)
+ (5 N) (
−
4 m)
=
−
13 J
.
004
(part 2 of 2) 10.0 points
Find the component of
vector
F
in the direction of
this displacement.
Correct answer:
−
2
.
54951 N.
Explanation:
W
=
vector
F
·
Δ
vectors
=
vextendsingle
vextendsingle
vextendsingle
vector
F
vextendsingle
vextendsingle
vextendsingle

Δ
vectors

cos
θ .
The component of
vector
F
in the direction of Δ
vectors
is
vextenddouble
vextenddouble
vextenddouble
vector
F
vextenddouble
vextenddouble
vextenddouble
cos
θ
=
W
bardbl
Δ
vectors
bardbl
,
so
F
cos
θ
=
W
radicalBig
s
2
x
+
s
2
y
+
s
2
z
=
1 J
radicalbig
(3 m)
2
+ (1 m)
2
+ (
−
4 m)
2
=
−
2
.
54951 N
.
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sousse (res2468) – HW 5 – Kleinman – (58225)
2
005
10.0 points
A block of mass m is pushed a horizontal dis
tance D from position A to position B, along
a horizontal plane with friction coefficient
μ
.
Then m is pushed from B to A.
If the force pushing m from A to B is
vector
P
,
and the force pushing m from B to A is
−
vector
P
,
what is the total work done by friction?
1.
2 (
P
−
μ m g
)
D
2.
2 (
μ m g
−
P
)
D
3.
+2
μ m g D
4.
0
5.
−
2
μ m g D
correct
Explanation:
From A to B, the work done by friction is
−
μ m g D
.
From B to A, the work done by
frition is
−
μ m g D
too since the work done
by kinetic friction is always negative. So the
total work done by friction is equal to the sum
of these two works, and we get
−
2
μ m g D
.
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 Spring '08
 Turner
 Physics, Energy, Force, Friction, Potential Energy, Correct Answer, wA

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