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