Unit 6 Review for Test€¦ · · 2016-08-23Unit 6 Review for Test SOLUTIONS ... 3. A 5 kg block...
Transcript of Unit 6 Review for Test€¦ · · 2016-08-23Unit 6 Review for Test SOLUTIONS ... 3. A 5 kg block...
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Unit 6 Review for Test
SOLUTIONS
Lyzinski Physics
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1. Draw out the full RUBBER STAMP that we start EVERY
incline plane problem with.
W
W||
W
||
FN
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2. A 4 kg block is sitting at rest on a 30o inclined plane.
Find the minimum coefficient of static friction that will keep the
block at rest. (TWO answers are expected: one should contain
only variables, and one should be a rounded number).
FN
Wperp
W||
Ff
577.)30tan(tancos
sin
cossin
)0(
coscos
cossin
||
||
||
mg
mg
mgmg
FW
mFWmaF
mgFFmgWF
mgWmgWmgW
f
fNet
NfN
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3. A 5 kg block is sitting on the frictionless inclined plane shown below.
A 20 kg weight is draped over the pulley and attached to the 5 kg
block via a cable. Find:
a) the acceleration of the system.
b) the tension in the rope.
c) the time that it will require for the block on the incline to travel
.5 m, assuming that it starts at rest.
50o
520
234.6254.29196
s
maa
MaFNet
5 20 20(9.8) = 196N 5(9.8)sin(50)
= 37.54N
a
(Continued on next)
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3. A 5 kg block is sitting on the frictionless inclined plane shown below.
A 20 kg weight is draped over the pulley and attached to the 5 kg
block via a cable. Find:
a) the acceleration of the system.
b) the tension in the rope.
c) the time that it will require for the block on the incline to travel
.5 m, assuming that it starts at rest.
sec4.get
oHarrison tMarvin Use
??
34.6
5.
0
2
1
t
t
s
ma
md
V
50o
520
20 a
T
20(9.8) = 196N
NTTN
maFNet
2.69)34.6(20196
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4. For the Atwood Machine at the right, find:
a) the acceleration of the system.
b) the tension in the rope.
288.5
58.92.39
s
ma
amaFNet
4 kg
1 kg
4 14(9.8) = 39.2N 1(9.8) = 9.8N
a
439.2N T
a
NT
TmaFNet
68.15
)88.5(42.39
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5. A box of mass m is being pushed up a o incline by a force of F.
The coefficient of kinetic friction between the block and the
incline is . Find the acceleration of the block, assuming that the
block does indeed move up the incline. (note: your answer
should contain only variables).
m
mgmgFa
mamgmgF
maWFFmaF
mgFFmgWF
mgWmgWmgW
push
push
fpushNet
NfN
sincos
sincos
coscos
cossin
||
||
FN
Wperp
W||
Ff
Fpush
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6. A 60 kg man is riding on an elevator while standing on a scale. A
50 kg chandelier is hanging from a metal wire in the ceiling
above the man’s head. Find:
a) the apparent weight of the man when the elevator is moving
upward with a constant acceleration of 3 m/s2.
NF
NmgmaWmaF
maWF
maF
N
N
N
net
768eight Apparent W
768)8.93(60
W=mg
FN = apparent weight
a
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6. A 60 kg man is riding on an elevator while standing on a scale.
A 50 kg chandelier is hanging from a metal wire in the ceiling
above the man’s head. Find:
b) the tension in the chandelier wire when the elevator is
accelerating downward at a constant acceleration of 2 m/s2.
NmamgmaWT
maTW
maFnet
390)28.9(50
W=mg
T
a
chandelier
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6. A 60 kg man is riding on an elevator while standing on a scale.
A 50 kg chandelier is hanging from a metal wire in the ceiling
above the man’s head. Find:
b) the minimum upward acceleration that the elevator must
constantly move at in order to snap the chandelier wire (which
is capable of withstanding 1000N of force).
2
2
2.10
)8.9(501000
)(
s
m
s
m
net
a
akgN
gammgmaWmaT
maWT
maF
W=mg
T
a
chandelier
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7. A man on skis stands at the top of a inclined hill. He begins to
descend. If the hill’s incline angle is 35o, and if the man travels
200 m down the hill in 10 sec, find the coefficient of friction
between the man’s skis and the snow. You may assume that
the man doesn’t push off with his ski poles, but rather simply
2.0)35cos()8.9(
4)35sin()8.9(
cos
sin
cossin
cossin
coscos
cossin
40sec,10,200
||
||
1 2
g
ag
agg
mamgmg
maFWmaF
mgFFmgWF
mgWmgWmgW
avtmd
fNet
NfN
s
m
FN
Wperp
W||
Ff
moves downhill due to the effects of gravity.
a
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8. Find the minimum angle needed to force a 50 kg block to slide down
an incline with a static frictional coefficient of 0.5.
o
o
f
NfN
mg
mg
mgmg
FW
mgFFmgWF
mgWmgWmgW
56.26
56.26)5(.tan)(tantan
cos
sin
cossin
on,acceleratian induce To
coscos
cossin
11
||
||
FN
Wperp
W||
Ff
a
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9. You weigh 600N and you are sitting in a space ship that blasts off.
The acceleration of the spaceship is 20 m/s2 upward.
a) Find the “apparent weight” that you will feel, in Newtons.
b) Find the “apparent weight” that you will feel, in g’s.
sg' 04.3600/356.824,1 sg' of #
356.824,1)8.920)(22.61(
)(
22.61600
NN
N
gammgmaF
mamgF
maF
kgmNmgW
N
N
net
W=mg
FN = apparent weight
a
person
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10. The system below uses rope that can withstand a
maximum tension of 35 0N. If the coefficient of
friction between the block and the table is 0.3, will the
rope snap or not? Make sure to fully explain your
answer.
2058.2
1002.88196490
s
ma
amaFNet
20 5020(9.8) = 196N 50(9.8) = 490N
a
30 kg
50
kg20
kg
30
Ff = (0.3)(30)(9.8) = 88.2N
(continued on next slide)
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10. The system below uses rope that can withstand a
maximum tension of 35 0N. If the coefficient of
friction between the block and the table is 0.3, will the
rope snap or not? Make sure to fully explain your
answer.
20 5020(9.8) = 196N 50(9.8) = 490N
a
50T 490N
a
!!!!!!!1.387
)058.2(50490
SNAPNT
TmaFNet
30 kg
50
kg20
kg
30
Ff = (0.3)(30)(9.8) = 88.2NCUT THE ROPE!!!!
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11. What coefficient of friction would be necessary to avoid a rope
“snappage” in problem #10?
30 kg
50
kg20
kg
Work backwards
50T 490N
a
2
max
8.250350490
350
50490
s
maaNN
NT
aTNmaFNet
(continued on next slide)
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11. What coefficient of friction would be necessary to avoid a rope
“snappage” in problem #10?
30 kg
50
kg20
kg
Work backwards
20 5020(9.8) = 196N 50(9.8) = 490N
047.)8.9)(30(
)8.2(100196490
)8.9)(30()8.2(100196490
)8.2(100)8.9)(30(196490
maFNet
30
Ff = ()(30)(9.8)From
previous slide
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12. Find the angle necessary to keep the system below at rest,
assuming that both inclines are frictionless.
20 50W||
o6.48)8.9(20
30sin)8.9(30sin
)8.9(20
30sin)8.9(30sin
30sin)8.9(30sin)8.9(20
1
20kg 30kg
30o
20 5020(9.8)sin
W||
20(9.8)sin(30)
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13. An Atwood machine has two blocks, of masses “m” and “M” respectively (where
M > m), being held at a height of “h” above the table top. The blocks are suddenly
released. Find …
a) the acceleration of the system in terms of “m”, “M”, and “g”.
b) the tension of the system in terms of “m”, “M”, and “g”.
c) the maximum height above the table that the less massive block will reach if
m = 5 kg, M = 12 kg, and h = 20 cm.
mM
gmMa
amMmgMg
)(
)(
m Mm Mmg Mg
a
M
Mg
T
a
cmcmcmcmH
myyyavv
vvahvv
mM
gmMa
sm
s
m
2.482.82020
082.)8.9(227.102
27.1)2)(.035.4(22
035.4512
)8.9)(512()(
22
1
2
2
2
2
2
2
1
2
2
2
h
mM
gmMMMgT
MaMgT
MaTMg
)(
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14. A block of mass “m” is placed on an incline whose angle of
inclination is “q”. If the block slides down the incline, and if the
coefficient of kinetic friction is “m”, find the acceleration of the
block in terms of m, , , and/or g.
agg
mamgmg
maFWmaF
mgFFmgWF
mgWmgWmgW
fNet
NfN
cossin
cossin
coscos
cossin
||
||FN
Wperp
W||
Ff
a
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15) The 10 kg block below is being pushed up the incline by a constant force of 150
Newtons. If the coefficient of kinetic friction on the incline is 0.2, find the …
a) normal force acting on the block.
b) speed that the block will attain after a 5 second push, assuming that
it starts at rest.
2
2
269.38)5(654.7)0(
654.7
10)918.110)(2(.)30sin()8.9)(20()10cos(150
sincos
918.110)10sin(150)30cos(*8.9*10
sincos
12 s
m
s
m
N
N
atvv
a
a
maFmgF
N
FmgF
FN
Wperp
W||
Ff
30o
a
10o
F
Fsin() Fcos()
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16) Two blocks (each of mass 70 kg) are connected together by a light string. They are
accelerated upward by a pulling force of 2000 Newtons. Find the tension in the
string while they are being pulled upward, assuming that they are both lifted off the
ground.
2486.4
14070702000
s
ma
agg
70
kg
70
kg
70g
70g
2000N
a
70
kg
70g
T
a
NT
gT
1000
)486.4(7070
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17) A military helicopter has a rope attached to the bottom of it, which is attached to a
platform on which a Hummer is sitting. The Hummer and the platform are resting
on a scale. The helicopter accelerates upward at a rate of 2.0 m/s2, and the rope
(which can handle a maximum tension of 30,000 Newtons) snaps at the exact
moment that the scale reads 1000N. Find the joint mass of the hummer and the
platform.
kgM
M
MM
MMF
12.627,2
8.11000,31
)2()8.9(000,30000,1
)2()8.9(1000
M
Mg
T
a = 2
1000N