Sheet for physic (1)

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School

Woburn Collegiate Institute**We aren't endorsed by this school

Course

PHYSICS SPH4U1

Subject

Law

Date

Dec 18, 2024

Pages

Uploaded by LieutenantBook16141

universal gravatation law Wc{?h‘( OG 5 O gicaion Fy, = mF, > nd (For moving objects) y\(\e&\'L Fn, Fg, Fa, F1on2, Ff, Fair Right+, up+, cw+ V1,v2,atdmtens ->Fnet=m->a Fa-Ff=ma ff=MkKTn G=6.67*10"N*m?/kg (constant) E.g. 1: A 15 kg box rests on a horizontal surface. a) What is the minimum horizontal force that is required to cause the box to begin to slide if the coefficient e of static friction is 0.35? b) What is the acceleration of the system if the person pushes the box with a force of 90 N if the coefficient of kinetic friction is 0.20? GXR | m= JSKa Ms= 035 Lp(bem:em of stadic fiiction. widude as force ?mfi: /WS Fr\/ Sowe MJ of 8row|'+/ o Fom= 035) (903) °, 4ne dorcc needS . - 51,45V - Fy L0 1o be aqrealer than b 51N Fr for Fa \ % b) ] @zawu\ Frei=ma > mufn o [ . fa - @ 150 Fa a0 - L3 (15349 - 1501 il > ?’8 .o the atcelerntiov éZi - s WS tml [rigni| 002 - g S Um [pa 1 <

4.3 Solving Friction Problems (part 2) E.g. 1: A &#kg sled experiences a 600 N force from a kid. If the coefficient of friction is 0.32, how far will the —— sled go in 4.0s? The sled starts at rest. h{-('.’.) V= Omls & N I J - Yi© Fa< (00N a: pg= 408 . 8] = A =~ =7 A N SRS BT AR % y—f—\ T o F\-F=500 1= 5(8.5bH) (07 A 2 — - Oy = 70 .4l %’Irb 600 /MKFN S0 fo=T ER‘\BW 1 600 ~ ou ey = 50 & 600 - (03 (50)(ag) = 0a o> e S led Accelerating (2)(3) EOO - ' 56 '? = SOQ (on:h-\ % oM rar Missing Variable Equation ] = SO o) v EJ{A:—%ZAN L‘ LFB a T G . — ¢ ; (e SO | = G bW, =a Ad v, = v, +aAl Z A —1 =1 e m Ie 1 B v, =v +2aé{77 IS " *** you must define a positive direction first *** % . %bq vj“jaiﬂ' M]a RigtCH) E.g. 2: A 65 kg toboggan is pulled by R32°U] force. If the coefficient of friction is 0.10, how long will it take to reach a speed of 5.2 m/s from rest? - S 0 ' = I ot | £,-as0N [R3a] B=GS M= 0.10 ?“"a“* P X - me 63K M- 1 e po= N 1A 1—;'\1 ¥ F(t) - Fo‘] a7 M L s Frevy = Fu-Fy =+ Foy 7 RS 0 = - () 5w (!l N e o Q,,(\v\«u\ O = FN— %1"’ ,gs g7 Fet = @ 0 = Fu - 45l 53% fay - e = 650 U31-53¢ = Hp HT-45.1538 =65 = Ak 31604 = 654 L 51y (451528 ) —“"G";’“ S % W ousasas s o fr= us.158 DCH] mls Vo= Vrobt 28wl - & 53= 0+ (3873) D1 Cege(l

E.g.2: Force of 65 N is needed to start a 8 kg box moving across a horizontal surface. a) Calculate the coefficient of static friction b) If the box continues to move with an acceleration of 1.4 m/s?, what is the coefficient of kinetic friction? G| Ty = 6N -y > > ca 1t (F /A"‘_il F{'ma% :/MS F'J b) N RBM > A\ f_;l‘rvw = /%S T N e, fa Hean " _ -};ne\ - Ma b -/%S fa — Fp = Mma ,» e (oeWiciend bS — MBS = 3(1-4) 0§ Staxic {riction ). (S - 78 Hme Il & - 78 4= [1-x=65 - 73.4m2 ~ 53-8 —--52.% -7%.d S =0. 680 AV S . t:p W 2] (OQQ’C i.d Q\Y\“ of lpnetic Aaction S 0.1,

E.g. 3: A force of 150 N pulls the 30 kg box to the right. If the coefficient of kinetic friction is 0.25, what is the horizontal acceleration of the box? - m‘im\] W‘:3OV\% = 0-25 P ) S > \ Fa A‘\.g [ Fy\c—\ = m-CbL fa — F£ = 200 150 — [045) [50)(A%) = 20a MK Is) — 73.35 = 20a ,o e 76.5 = 204 pccelerviion 765 - 4 S 3nk” fﬂ‘a@ = 255 = o imlgZ I E.g. 1:A 20.0 kg box is pushed horizontally across a floor with a force of 50.0 N to the right. The coefficient of static friction for the two surfaces is 0.45 and the coefficient of kinetic friction is 0.40. Determine all the forces acting on the box if.....,Pfiaml a) Itis moving b) Itis stopped R | m= QOMb /,{/9 - 0&5 Mo = 0-¢0 AS | Fq = 50.0N [ﬂ‘ah“] bj |§g:/lfis F;\\)_ }F’S = (0.U5) (0Y48) e = g"‘o(ce - N Q) TP 7, 2= dban Fe- 0\ GoV ) ) [ ks %fl\l‘ﬂfij Fu= 7% -4 N & the kinetic Fu = Lriction 15 78N (ed] Fr= 74N (Lt ) s e 1 E.g. 2: Determine the coefficient of friction if it takes a 62 N force to get a 22 kg box to move across a floor. G aav;é - s _7 ?S = éa E\Q-Hj Mol & - 7/%5:0'987 2 FS =l sFN . o /“ 5 the eficient b = s (N(25) 0¥ faddion b 15 0.9, (a e

E.g. A 45 kg toboggan is pulled up a hill (15° to the horizontal) by a force of 310 N 12° above the surface of a gi]ll.slf it starts from rest, how long will it take to make it up the 6.2 m hill. Assume the coefficient of friction is RN A Fa _ 50N [12°aboe Surfue of kil ] = qfikg, 0-5 M - 2m /% . Up fomp C+> (0\ Yoy AtS | g - M FL USIGERTN 6 % 0 = wsamy = (Us)(a 7) [1nt0 = 4N Cdoun] ronp | ¥ y < v : l}:a\\ = [45)(01.8\g;h(;5°> s = 14159 [don cung] Fau = 510 s (122) oWt 0f remp (1) w = 1498855 N [uprone | an, P + R+ Fu Far = 510 sin (2°) = —UR5AT3+00. 03¢ 3 Fy Fay T Db 034N [putof Y“W‘ﬂ 0 =319.9%9 + Fy Fne-\u = Mo A=ty MV Fau — Fe-Faa = 4Sa §ag. %55 — /n5) ( 310.929) — 1y 137 = YSq