DYNAMICS REVIEW SOLUTIONS
I f'TFA 78N TEA
120N
FNET ma
Fg Fan Fatma Faa
M
120 789400.047 FRIGHT
2 NEWTONS FIRST LAWTHE PASSENGER HAS INERTIA UNTILA FORCE FROM THE BUS ACTS ON
THE PASSENGER HE SHE CONTINUESTO MOVE AT A CONSTANT
VELOCITY FORWARD
AS THE Bus Slows DownTHE PASSENGER DOES NOT
3 KINEMATICS 0
Vi O Vf tat
Uf 16 Is Vf at1 5.05
a Ifa t
1653.272
FNET ma
1200 3.2
3840N IN THE
DIRECTION OF THE
FINAL VELOCITY
4 A FB on A BALL
ID D FA on B
ASTRONAUT
NEWTON'S THIRD LAW
FB on A FA on BM BAB
f 4 3.3
7.92 N DIRECTION
7 9 N LEFT OPPOSITE OF
FA on B
b Fine ma
a FNETT7.9 2
86
0.092 Fz LEFT
DIRECTION OPPOSITE OF FA on B
5 a FN the 1 ma 4 stanFn Fg 0 SPEED
D Fn FgmgLeg62719 8
608 Nb f Fn ma 610N
D Fn Fg ma
Fn mg _ma
tfg Fn matingm atg
62 1.8 9.8
719 N
720N
c
qFN Fn T ma
D Fn Fg ma
Fn mg _ma
Fn a'Igg9 621 2.5 9.8
453N450N
GFn Fg UP Down
Ff Fa mgBALANCED
tMaEEo E
Fg Fa Ff D
WHEN CONSIDERINGL
FA FfTHE SCENARIO WHERE
µsFNAN OBJECT JUSTMsmgSTARTS TO MOVE
USE Ms AND a O Ms FImg
381 9.8
0.46
b
IFN Fn Fg UP Down
F g Fa mgBALANCED
FNET Ma E tartFg FA Ff 0 VELOCITY
FA FfMENMing
Mk _Iamg
120381 9.8
0.32
7gen
Fn FYI EisF
f NET _ma
IFg Ff ma
µ For ma
mYg m a
ng a
a 0.62 9.8
6.076 Fr
KINEMATIC S
a 6.076 Fr Vf Vi Zad
Vi 185 O Vi't 2 ad
Vf O dD
za
1822C6.076
27M
8 a IFN Fn _Fg vP DownBALANCED
D my
Te Ff µFnp mg0.287 4.87 9.8
13 N LEFTDIRECTION OPPOSITE
OF ITS VELOCITY
I II Ono n
FIND FNTD
F 1 Ma
Fg Fn Fg ma
Fr ng ma
Fn mating
m at g4.811.61 9.8
Ff _µFn 54,72 N
0.28 54.72
15N LEFT DIRECTION
OPPOSITE OF ITS VELOCITY
FRICTION INCREASES AS FN INCREASESc
GINCONSIDER ONLYUP DOWN MOTION to
ID FIND FN
te te ma
Fn Fg ma
Fr ng ma
Fn mating
m at g
Ff µFn4.8C 1.61 9.8
10.287 39.3639.36N
11N LEFT DIRECTION
OPPOSITE OF ITS VELOCITY
FRICTION DECREASES AS FN DECREASES
9 Ff Ff Five _ma a 0
NOT
f f Fftff Fg O FALLINGIn Fa D FN FA
2Ff Fg o
2 MEN mg 0If2MFA m.fi
m9ZM
3.211 9.8210.53
30 N