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Physics
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Newton's Law of Motion:
5 He Avec 6 Moh
@ sme archon v/ mobo
WrematiCs — nostron oy
1. Superman must stop a tran in 150 m to keep 120 kmvh it from hitting a stalled ear on the
tracks. Ifthe train's mass is 3.6 10° kg, how much force must he exert?
om aw”
a
= 7a a
= —_—
Yer 0b Spe
kK ——_ 19m —_4
Kamematicg +
recall 20 Nps Ye at (se, >
© s~ wt +tat Ce, ay
ONPy Yr + as %,s)
Fu = a
-Fp = ma
mar
te = Go tn®\3 73)
Re
1,333.68 KN
2A wh stands on a bathroom scale in a
Of the elevator, and find the direction of acceleration.
Wa Wes ma
Ne 07 = yaw
ae 78+ 0.95422)
awn ware)
‘When the elevator begins
to mdve, the scale briefly reads only 0.75 of her regular weight. Calculate the acceleration
W ; Found. diewawoed ewstoy,
z A> me
KimeCs — atin w/ Fores
Nox Bo Fit Jem fhe
fir (em 36005
No= (120= 3.6)
No= 33.33 m/s
= Yo 4 2aS
OF 33.33% 4 2(@C{d)
™m me cclsity
2 >
Pn
one
=
Fopine. upward — ewoty
ar me
MWi- Ws ma
O-Fibg ~ ig = ha
a= 0.4531) — 7.31
=O m
a Cars
ach erting, ss [ s
town ware] tee
Gurmg ayasnp tion >)
3. Two crates, of mass 65 kg and 125 kg, are in contact and at rest on a horizontal surface. A
650-N force is exerted on the 65-kg crate. If the coefficient of kinetic friction is.0.18, ~ MK
calculate (a) the acceleration of the system, and (b) the force that each crate exerts on the
‘other. (c) Repeat with the crates reversed.
ay =0
N= We 6543) = GAGS N
fhe N= (28@3) = 2225 W
Bufo seme wy Ah _ Leek - Ne!
a
Cfeugerticl > Sng wy 5A
ae reefer
Circular Motion
2 y/
Radial (centripetal) Acceleration: ag == ae 2m
OX T
Centripetal force: ¥ Ry = mag
Lindi
Le frees actig along arbipet, | Fate,
He Botue os
© ene w/awtion 5 rowarel
© oppsite > owhwerel ©
by
fren]
Sony free
4. Aen sven GTA ow of de by moving nan ag eT} Y
What isthe plane's acceleration?
lane's aveeT ati 1
2
yt _ [ten 20]
ee ~ =] 538.0 2!
5,200
5. A horizontal force of 310 N is exerted on a 2.0-kg ball as it rotates (at anm’s length)
uniformly in a horizontal circle of radius 0.90 m. Caleulate the speed ofthe ball
0am —4
- -
on
2
< Vv
a = 2 )
Comp
[ye nary
6. A0.55-kg ball, attached to the end ofa horizontal cord, is revolved ina circle of radius 1.3
‘mona fitionless horizontal surface. I the cord will break when the tension init exceeds 2
75N, what is the maximum speed the ball eanhave?” (game wy/ HS) aF-m(£)
7. 4.0.90 kg body attached to a cord is whirled in a vertical circle of radius 2.50 m. What
migigum ust it have at the top of the circle so as not to depart from the circular
path?
t
T CF amas mt
£
20m | |) aor ee a IS am fin WitTad = m(E)
P= (CwtTend) (&)
LN VG Teed EY
itn Lntable.
Tam = 0 >
[organ + 0] (22) = [uae w/e
8. Acarat the Indianapolis Sifaccelerates uniformly from the pitaea, going from rest 9 =
a semicircular are with a radius of 220 m. Determine the and zadial acceleration
of the car when it is halfway through the arc, assuming constant tangential acceleration.
s= SQM) = lor
Nts Not 4 245
2
Gre “. eas (.2626 4 ] 2
lesa, Gan Shift IF
Gales Garo: Wht 7 #34 Oores= Naz tax’ = L375 ofr
Gh: Crealue 439, Shy
n=
6= Ga xn! " the
Lawof Universal Gravitation, Gok Fey me woes
Gravitational Force: F =@f22 Ea
r= Im
9. Two objects attract each other gravitationally with a force of 2.5 x 10" N when they are 3
0.25 m apart. Their total mass is 4.00 kg. Find their individual masses. fi= 220 NN
10. Calculate the acceleration due to gravity on the Moon, which has radius 1.74 x 10° mand
mass 7.35 x 10” kg.
= HM
goth fer oC SE
G-* m= dm= 34 Key
M= Ba40l ky > me dam= 90544
fe
= = 6 "i
ae = (la
i Y
2
A a
Me = S94
< = |\¢7ai¢ m
Jecrth Coe aie
(B- FAs”) ed)
: Gwars= cmb GBI? ~[s m
ce
iG 3539.5 xl”) —
eek ?
O29 O
fee low fooke
Ui Ace
Morte Fed % FIN srbhy
Work and Energy
Work W = F xd Werk. = fet % & [| A = fruslaty
Kinetic Energy: KE = mv?
‘Work-Energy PrinciplesWnat=DKE \
Potential Energy: PE = mgh
Potential Energy of Elastic Spring: PEs» = kx?
‘\Gonservationof Mechanical Encruy: KE, + PE, = KE, + PE,
cowrkirt, speed, AKE-O cmerbnt, speed , OKE~O
11. How much work gid the Atovers do (horizontally) pushing a 46.0-kg crate 10.3 m across a
rough floor Githout BeéCteratiOy if the effective coefficient of friction was 0.50?
Movers
—? | 4,
——
qo feaew
N=w
({<—_—— 0.3m —— (Fa
Ntet= Mis - Wreg = 0 Wanrs Gala (w.3) *e
7 ee tee 5 Veena |
= Hare [£212] =0
12. What is nC nimum wo ded to push a 950-kg car 710 m up along a 9.0° incline?
Ignore friction:
Wht 0d
ame = Wee ween 20“
A! re hh
i ?- Wan — FGI) GidEmz") =6
thes) 1099 —
No= 95
13. How much work must be done to. vee .925-kg car traveling at 95 km kav/h?
Wovt= = Joe a)
z =
worke—eneray et (me[ 0-2 ] = |- 322.07 7
Predple 36 L,
14, By how much does the gravitational potential energy of a 54-ky pole vaulter change if her
center of mass rises about 4.0-m during the jump?
w2— PEs onl = Sef 4.B1)(4) = 2118-16 lm
{ —<—$$——
fam . = PE tng = SU ABIIL4) = 21s Te Nom
1 oo
a
15. A spring has a spring constant k of 88 sa. How much must this spring be compressed to
store 45.0 J of potential energy?
ase £08) ol
X= Lollm \
‘A spring with k = 83 Nim hangs vertically next to a ruler. The end of the spring is next to
the 15-cm mark on the ruler, Ifa 2.5-kg mass is now attached to the end of the spring, and.
the mass is allowed to fall_where will the end of the spring line up with the ruler marks
when the mass4s.at its lowest position’
Gonshretehed ) H-. &
+o —
ch ty 1h HA
Gir) 0+ 2.6(4. a1) (ais) 10> ot zs(anexaisi)
(Sem
\oro £49097
«fate yee Ol m
z ee
) Reodng ;
© e dere
leeds,
= OIS+4 0.54]
= Ot
\2=0 Neo
