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Physics 20 - Lesson 18Dynamics – Pulleys and Systems
Possible 125 / 71
1)
/4
2)
/6
3)
/7
Tension – I chose the 3.0 kg mass
+
_
Dr. Ron Licht Answer Key 18 - 1 www.structuredindependentlearning.com
●
FT
5
3
acceleration
using the 5.0 kg block
3.05.0
calculate acceleration first
1540
100
25
nF
100gF
fFr
+
25gFr
2
25
25
25
25 (9.81 )
245.25
g
mg s
g
F mg
F kg
F N
rr
2
100
0.20(100 )(9.81 )
196.2
N g
f
mf s
f
F F mg
F mg
F kg
F N
rr
Acceleration125 kg
25gFr
fFr
Tension100 kg
TFr
fFr
7Fr 7.0kg
Tension
TFr
4)
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5)
/11
Dr. Ron Licht Answer Key 18 - 2 www.structuredindependentlearning.com
+
B
A
Acceleration
12.0kg
●
●
●
●
6)
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7)
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Dr. Ron Licht Answer Key 18 - 3 www.structuredindependentlearning.com
c)
b)
a)
The force on each girl will be equal (360N) but opposite in direction
Acceleration
30kg
(–) (+)
= +60 N20 kg10 kg
8a)
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b)
/5
c)
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Dr. Ron Licht Answer Key 18 - 4 www.structuredindependentlearning.com
a negative net force indicates that the men do not move in the positive direction, therefore a = 0
FgC
883N 932N
= +932N
= –883N●
BTAT
16 kg 24 kg 20 kg
135 N
A BgB BF m g
r
– +
A
B
25.0 (9.81 )
49.05
mup s
up
F kg
F N
rr
2(9.81 ) mdown B sF mr
+
–
Bonus1)
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Bonus 2)
/5
Bonus 3)
/8
Bonus 4)
Dr. Ron Licht Answer Key 18 - 5 www.structuredindependentlearning.com
a) find mass B
a)
b) TA may be calculated using the 16 kg girl
TB may be calculated by combining the mass of the 16 kg girl and the 24 kg girl
b) find tension – use mass A
6.0 kg
••
(0.5) 4.905NF N(1.0 0.5)
(1.0 0.5)
9.81 4.905
14.715
N
N
F N N
F N
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Bonus 5)
/8
Dr. Ron Licht Answer Key 18 - 6 www.structuredindependentlearning.com
The inertia of the 0.5 kg block will be overcome when the acceleration of the system exceeds the friction force between the two blocks.
Using the acceleration we can calculate the applied force.
Bonus 6)
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Dr. Ron Licht Answer Key 18 - 7 www.structuredindependentlearning.com
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