Physics 20 - Lesson 18
Dynamics – Pulleys and Systems
Possible 125 / 71
1)
Fnet  ma
15
40
Fg15  (m1  m2 )a
/4
147.15 N  (40kg  15kg )a
a  2.68 m s 2
calculate acceleration first
2)
5.0
3.0
Fnet  ma
29.43 N  (8.0kg ) a
/6
a  3.68 m s2
using the 5.0 kg block
FT
Fnet  ma
FT  5.0kg ( 3.68 m s2 )
FT  18.4 N
acceleration
3)
Fnet  ma
Fdown  Fup  (m1  m2 )a
3
49.05 N  (29.43 N )  (3.0kg  5.0kg ) a
/7
19.62  8.0a
5
a  2.45 m s2
0
Tension – I chose the 3.0 kg mass
Fnet  ma
+
FT  Fg  ma
FT
FT  (29.43N )  3.0kg (2.45 m s2 )
●
FT  36.8 N
_
Dr. Ron Licht
Answer Key 18 - 1
www.structuredindependentlearning.com
+
Fn
4)
Ff
Acceleration
125 kg
●
100
Fg 25
Ff
/12
Fg100
25
Fnet  Fg 25  Ff
FN  Fg100  mg
Ff   mg
Fnet  245.25 N  (196.2 N )
Fg 25
Fnet  49.05 N
Ff  0.20(100kg )(9.81 m s2 )
Fg 25  mg
Ff  196.2 N
Fg 25  25kg (9.81 m s2 )
a
Fnet 49.05 N

m
125kg
a  0.3924 m s2
Fg 25  245.25 N
Tension
100 kg
●
Ff
FT
Fnet  FT  Ff
FT  Fnet  Ff
FT  ma  Ff
FT  100kg (0.3924 m s2 )  ( 196.2 N )
FT  235.4 N
Fnet  F7  F5
5)
+
Acceleration
Fnet  68.67 N  (49.05 N )
Fnet  19.62 N
/11
●
12.0kg
a
a  1.635 m s 2
A
B
Fnet  F7  FT
Tension
FT
FT  Fnet  F7
FT  ma  F7
7.0kg ●
FT  7.0kg ( 1.635 m s2 )  68.67 N
F7
Dr. Ron Licht
Fnet
19.62 N

m (7.0kg  5.0kg )
Answer Key 18 - 2
FT  57.2 N
www.structuredindependentlearning.com
The force on each girl will be equal
(360N) but opposite in direction
6)
a)
a1 
/12
b)
Fnet 360 N

m1
40kg
a2 
Fnet 360 N

m2
60kg
a1  9.0 m s2
a1  6.0 m s2
v2  v1  a t
v2  0  6.0 m s2 (0.10 s)
v2  0  9.0 m s2 (0.10 s)
v2  0.60m / s
v2  0.90m / s
c)
1
a t 2
2
1
d  (9.0 m s 2 )(0.10 s) 2
2
d 
d 
1
(6.0 m s2 )(0.10s) 2
2
d  0.030m
d  0.045m
7)
(–)
10 kg
20 kg
(+)
= +60 N
FN 10  Fg10  10kg (9.81 m s 2 )
FN 20  Fg 20  20kg (9.81 m s 2 )
FN 10  98.1N
FN 20  196.2 N
Ff 10    FN 10  0.15(98.1N ) Ff 20    FN  0.15(196.2 N )
/9
Ff 10  14.715 N
Ff 20  29.43N
Fnet  FA  Ff 10  Ff 20
Acceleration
Fnet  60 N  (14.715 N )  (29.43 N )
30kg
Fnet  15.855 N
a
Fnet 15.855 N

m
30kg
a  0.53 m s 2
Dr. Ron Licht
Answer Key 18 - 3
www.structuredindependentlearning.com
8a)
Fnet  FgC  FgL
= –883N
Fnet  932 N  (883 N )
●
Fnet  49 N
= +932N
a
/5
883N
Fnet
49 N

mT
95kg  90kg
a  0.265 m s 2
932N
b)
/5
Fnet  FgC  F f
Fnet  932 N  (794.7 N )
Fnet  137.3N
Ff   FNL
Ff  0.90(883N )
Ff  794.7 N
a
Fnet
137.3 N

m T (95kg  90kg )
a  0.742 m s 2
c)
FgLy  cos 60 (883 N )
Fnet  FgLC  FgLx  Ff
FgLy  441.5 N
Fnet  932 N  (764.7 N )  (176.6 N )
FgLx  sin 60 (883 N )
a negative net force indicates that
the men do not move in the positive
direction, therefore a = 0
FgLx  764.7 N
/6
Fnet  9.3N
FgC
Ff   FN
Ff  (0.40)(441.5 N )
Ff  176.6 N
Dr. Ron Licht
Answer Key 18 - 4
www.structuredindependentlearning.com
Bonus
1)
16 kg 24 kg 20 kg
F net  135 (no fricton)
a)
Fnet
135 N

m 16kg  24kg  20kg
a  2.25 m s2
a
135 N
/9
TA
TB
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
Fnet  ma
TA  ma
Fnet  ma
TA  16kg (2.25 m s 2 )
TB  ma
TA  36 N
TB  (16kg  24kg )(2.25 m s 2 )
TB  90 N
Bonus
2)
Fnet  ma
FgB  (mA  mB )a
FgB  mB g
A
 mB g  (mA  mB )a
B
9.81 m s2 mB  (170kg  mB )( 2.5 m s2 )
/5
9.81mB  2.5(170)  2.5mB
–
7.31mB  2.5(170)
+
mB  58.1kg
Bonus
3)
a) find mass B
Fnet  ma
Fup  5.0kg (9.81
m
s2
)
Fup  49.05 N
/8
+
A
–
B
Fup  Fdown  (mA  mB )a
49.05 N  (mB g )  (5.0  mB )(4.0 m s 2 )
49.05  9.81mB  20  4.0mB
69.05  5.81mB
mB  11.9kg
Fdown  mB (9.81 m s2 )
b) find tension – use mass A
FgA  49.05 N
Fnet  ma
FgA  FT  ma
49.05  FT  5.0kg (4.0 m s 2 )
FT
Dr. Ron Licht
FT  69.05 N
Answer Key 18 - 5
www.structuredindependentlearning.com
Bonus
4)
6.0 kg
FNET  ma
Fg  Ff  Fx  ma
29.43N  (5.097 N )  (14.715 N )  6.0kga
9.62  6.0a
9.62
a
6.0
Fx
/8
Fx   Fg sin 
a  1.60 m s2
Fx  29.43sin(30)
Fx  14.715 N
Ff    FN    Fg cos 
Ff  0.20(29.43 N )(cos 30)
Ff  5.097 N
Bonus
5)
FN (1.00.5)  9.81N  4.905 N
FN (0.5)  4.905N
FN (1.00.5)  14.715 N
•
/8
•
Ff 0.5  0.5 FN (0.5)
Ff 0.5  0.35(4.905 N )
Ff (1.0 0.5)  (1.00.5) FN (1.00.5)
Ff 0.5  1.72 N
Ff (1.0 0.5)  0.20(14.715)
Ff (1.0 0.5)  2.94 N
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.
F f 0.5  ma
a
F f 0.5
m
1.72 N
a
0.5kg
a  3.43 m s 2
Using the acceleration we can
calculate the applied force.
Fnet  ma
FA  Ff (1.0 0.5)  ma
FA  ma  Ff (1.0 0.5)
FA  1.5kg (3.43 m s2 )  (2.94 N )
FA  8.09 N
Dr. Ron Licht
Answer Key 18 - 6
www.structuredindependentlearning.com
Bonus
6)
–
Ff
v i  20 m s
Fnet  Ff  Fx
vf  0
ma  (1.703m)  (0.966m)
ma
  2.669m

m
a  2.669 s2
a?
10
/ 10
Fx  Fg sin 
d?
Fx
Fx  mg sin(10)
v f2  vi2  2ad
Fx  m(9.81 m s 2 ) sin(10)
d
v f2  vi2
2a
0  (20 m s ) 2
d
2(2.669 m s 2 )
Fx  1.703m
Ff   FN
d  75 m
Ff   Fg cos 
Ff   mg cos 
Ff  0.10(m)(9.81 m s2 )(cos10)
Ff  0.966m
Dr. Ron Licht
Answer Key 18 - 7
www.structuredindependentlearning.com