markowitz (am45362) – Practice Questions for Exam 1 – distler – (56295)
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001
10.0points
A freight train has a mass of 2
.
5
×
10
7
kg.
If the locomotive can exert a constant pull
of 5
.
4
×
10
5
N, how long would it take to
increase the speed of the train from rest to
89
.
2 km
/
h? Disregard friction.
Correct answer: 1147
.
12 s.
Explanation:
Let :
m
= 2
.
5
×
10
7
kg
F
net
= 5
.
4
×
10
5
N
v
f
= 89
.
2 km
/
h
vector
F
net
=
summationdisplay
vector
F
=
mvectora
a
=
F
net
m
=
5
.
4
×
10
5
N
2
.
5
×
10
7
kg
= 0
.
0216 m
/
s
2
.
Since
v
i
= 0 m/s,
v
f
=
a t
Δ
t
=
v
f
a
=
89
.
2 km
/
h
0
.
0216 m
/
s
2
·
1000 m
1 km
·
1 h
3600 s
=
1147
.
12 s
.
002
10.0points
The horizontal surface on which the objects
slide is frictionless.
The acceleration of gravity is 9
.
8 m
/
s
2
.
1 kg
9 kg
4 kg
F
ℓ
F
r
If
F
ℓ
= 20 N and
F
r
= 6 N, what is the
magnitude of the force exerted on the block
with mass 9 kg by the block with mass 4 kg?
Correct answer: 10 N.
Explanation:
m
1
m
2
m
3
F
ℓ
F
r
Given :
vector
F
ℓ
= +20 N ˆ
ı ,
vector
F
r
=
−
6 N ˆ
ı ,
m
1
= 1 kg
,
m
2
= 9 kg
,
m
3
= 4 kg
,
and
g
= 9
.
8 m
/
s
2
.
Note:
F
is acting on the combined mass
of the three blocks, resulting in a common
acceleration after accounting for friction.
m
1
F
21
F
ℓ
m
2
F
32
F
12
m
3
F
r
F
23
Let
F
ℓ
, F
r
, F
32
represent the force exerted
on the system from the right, from the left,
and the force exerted on
m
2
by
m
3
, respec
tively.
Note:
vector
F
21
=
−
vector
F
12
and
vector
F
32
=
−
vector
F
23
, and
bardbl
vector
F
21
bardbl
=
bardbl
vector
F
12
bardbl
and
bardbl
vector
F
32
bardbl
=
bardbl
vector
F
23
bardbl
, where
bardbl
vector
F
bardbl ≡
F
is the magnitude of
vector
F
.
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 Spring '08
 Kaplunovsky
 Physics, Force, Friction, Mass, Velocity, kg

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