COURSE TARGETS AP PHYSICS TEST SCORES World SHS

2011 AP PHYSICS TEST SCORES World SHS 2011 AP PHYSICS TEST SCORES World SHS COURSE TARGETS be ble to stte, nd understnd the mening of, Newton's 3 lws of motion. be ble to pply Newton's lws to simple situtions in one nd two dimensions. be ble to drw free-body digrms. know the difference between weight nd mss. be ble to pply Newton's lws on inclined surfces. be ble to perform force nlysis in situtions involving both sttic nd kinetic friction. NEWTON'S LAWS MISCONCEPTIONS Action-rection forces ct on the sme body. There is no connection between Newton's Lws nd kinemtics. Velocity is force. The product of mss nd ccelertion, m, is force. Friction cn't ct in the direction of motion. The norml force on n object is equl to the weight of the object by the 3rd lw. The norml force on n object lwys equls the weight of the object. Equilibrium mens tht ll the forces on n object re equl. Equilibrium is consequence of the 3rd lw. Only nimte things (people, nimls) exert forces; pssive ones (tbles, floors) do not exert forces. Once n object is moving, hevier objects push more thn lighter ones. Newton's 3rd lw cn be overcome by motion (such s by jerking motion). A force pplied by, sy hnd, still cts on n object fter the object leves the hnd. 1

The spce shuttle Endevor lifts off All of Newton s lws of motion re exhibited during this lift-off. Do objects fll t sme velocity in vcuum So they hit with t the sme time over the sme distnce. Do they hve the sme velocity? NASA So then do they hve the sme force of grvity on them? http://www.youtube.com/wtch?v=_mcc- 68LyZM&feture=relted&sfety_mode=true&persist_sfety_mode=1&sfe=ctive NEWTON S LAWS The first lw defines inertil reference frmes nd the motion of n object with blnced forces. It does not hold if ccelerting frme. The second lw describes the reltionship between force, mss nd ccelertion. The third lw describes the forces of interction. A chemist might be sked to weigh out 200 g of certin element. WRONG Also, I her bout 10-kg lod s if it ws Weight. WRONG The kilogrm is mss - never force - nd it doesn t hve direction or vry with grvity. Weight is Force The unit of Force (weight) is the Newton (kg m/s2) 1Kg of pltinum-iridium on Erth = 1Kg of pltinum-iridium on the Moon (MASS) But it Weighs 8.9 N on the Erth nd 1.6 N on the moon. (FORCE) IDEAS OF INERTIA Glileo s contribution to the science of motion. Glileo's dynmics the rrow continues to fly through the ir becuse of the lw of inerti. A block of wood on tble stopped sliding once the pplied force ws removed becuse of frictionl forces tht Aristotle hd filed to nlyze correctly. WOOD ON WOOD DEMO Criticl Lerning Point!! Ancient Criderin Lw: In Physics, the use of Newton s second lw nd mny other pplictions mkes it bsolutely necessry to distinguish between mss nd weight. Use the correct units (or suffer in the cdemic gllows!) Metric SI units: Mss is in kg Weight is in N. 2

A freeflling meteor hs mss of 10.0 kg, FIND THE WEIGHT (THE FORCE OF GRAVITY NOT THE INERTIA) Newton s First Lw: Object t rest remins t rest, nd object in motion remins in motion with constnt velocity unless it experiences net externl force. When net externl force is zero, the ccelertion is zero w = m w = 10.0 kg (9.80 m/s 2 ) w = 98.0 kg m/s 2 w = 98.0 N Units of Force -- Force is vector Newton (N) Cge em in 2 seconds Put tiger cged here BALANCED FORCES NEWTON S FIRST LAW In USA - Pound 1 N = 0.225 lb 1 N = qurter pounder (no cheese) Object t rest V=0m/s A = 0 m/s2 Object in motion V=0m/s A = 0 m/s2 Weight force exerted by grvity on object s ms Weight cn chnge if grvity chnges Inerti is the resistnce of n object to chnge in it s tte of motion. Moon - 1/6 of your weight on erth. Inerti depends solely on mss Ask them if they feel the force of grvity on them Wht if you were to pull the floor, would they be weightless (AKA grvity stops) Hve them stnd up. Do they still feel grvity? Why is it on their feet? UNBALANCED FORCES Accel / Decel V=0m/s A = 0 m/s2 NEWTON S SECOND LAW Displcement V=0m/s A = 0 m/s2 Unblnced Forces will cuse ccelertion 3

No Motion BALANCED OR UNBALANCED? BALANCED Free Body Digrms F frictionless Motion UNBALANCED 1. FBDs lwys originte from the center of mss 2. FB rrow length reflects mgnitude nd direction SOMEWHAT BALANCED FN F A F N No Motion (For Now) W B WR W R F f FREE BODY DIAGRAMS NEWTON S SECOND LAW SO WHAT IS THIS FORCE? The bility to ccelerte mss with inerti It is defined s push or pull, but there re 5 types of forces. Compression, Tension, Sheer, Torsion, Bend, Contct forces: Objects in contct exert forces. Convention: F,b mens the force cting on by b. The Crider Mn contct Force Ctegories Norml f N surfce friction Ropes/rods f T Sttic f s Kinetic f k F hed,thumb You So F hed,thumb mens the force on the hed due to the thumb. Non-contct Grvity F g Electric F E Mgnetic F B 4

Unblnced forces BLOW GUN EXAMPLE An unblnced force of 20N gives stone n ccelertion of 4.0m/s 2 wht is the mss? A Force hs mgnitude & direction (vector). Adding forces is like dding vectors. They cn be up to 3-Dimensionl F 1 F NET = m Think of 2 forces tht don t fit in the previous 5 types F 1 F 2 F NET F 2 Accelertions chnge velocities Forces cuse ccelertions (dependent on mss) Unblnced forces An unblnced force gives 2.0kg mss n ccelertion of 5.0m/s2 wht is the force? http://www.youtube.com/wtch?v=rzd62ohdti&feture=relted 5

Newton s Second Lw For ny object, F NET = m Accelertion of n object is proportionl to the net force F NET cting on it. The constnt of proportionlity is clled mss, denoted m. The mss of n object is constnt property of tht object, nd is independent of externl influences. http://www.youtube.com/wtch?v=kdp1tiuszw8&sfe=ctive Would objects freely flling on the moon lso hve equl ccelertion? Yes, but it would be 1/6th of the grvittionl ccelertion on erth. If the mss of n object is doubled, why doesn t the greter inerti ffect slow grvity's ccelertion? When the mss is doubled, the weight is doubled. Grvittionl force depends upon mss. 6

NORMAL FORCE Different from Newton s 3rd http://www.youtube.com/wtch?v=obdlzrjr5g Strt the clss by hving ech student jump up nd down on scle Probe: We sid lst week tht your bottom gets tired fter 4 hours of the SAT? Led them to forces--- We sid lst week tht your bottom gets tired fter 4 hours of the SAT? Led them to Norml Force--- Give me some exmples of norml force? Could bthroom scle work on the moon, how bout in orbit on the spce shuttle? Speking of tht, how would spcemn know if he ws gining weight? Wow if you lived in spce, tht could be slight problem. If you st on bthroom scle going through loop would you weight more? Why does the scle get In the elevtor lb you Would you wnt n elevtor tht llowed you to ccelerte t 9.8m/s (T&T) NEWTONS THIRD LAW 7

Now sme problem, sme 250 N force nd cr, but the mss of the cr hs twice the mss F =m() F 2m F = m = 0.050 m/s 2 Accelertion is only ½ of previous problem. Which of Newton s Lws is represented by the grphs below? F 1 2 A monkey with mss of 1.2 kg slides down vine. The vine exerts n upwrd force on the monkey of 0.06 N. Show mthemticlly wht the ctul ccelertion of the monkey is. Newton s 2 nd Lw slope of grph 1 = constnt mss tht force is pplied to constnt force pplied to incresing mss in grph 2 definition of inertil mss resistnce to ccelertion m A 2500 kg cr is pushed with 250 N force, wht is the ccelertion cting on the cr? Does this men tht upwrds? F =m() F m F = m Describe how we could pproch finding the Monkeys velocity fter sliding down the 5m rope = 0.10 m/s 2 8

This imge cnnot currently be displyed. BLOWGUN WHY DO WE HAVE SO MANY SECTIONS? DOES THE DISTANCE FROM THE TARGET MATTER TO THE VELOCITY IT HITS AT? (ssume no wind resistnce) Lerning Point RATIONALIZE HOW FORCE CAN BE A VECTOR IF MASS IS A SCALAR? An elephnt gun cn fire 500g drt out of the Brrel t 200 m/s. The length of the brrel is 1.2m. Wht force is on the bullet? Foiled By A box of mss m =10 kg is t rest on sheet of ice (frictionless). You pull on rope tht is ttched to the box nd pulled t n ngle of = 30 o bove horizontl with tension T = 40 N. Mx the dog (25 kg) pulls with ginst you with 30 N of force. 1. Does the box lift up s you pull it? 2. Wht is the box s velocity fter five minutes? T m physics FORCE IS A VECTOR Sme s with Vectors F R = F 1 + F 2 Tn θ = F 2 / F 1 Force Problem Solving STEP 1: Sketch the sitution!! Drw Free Body digrm for EACH body in problem & drw ALL forces cting on it. Prt of your grde on exm & quiz problems! F N T f MAX Fg 9

STEP 2: Resolve the forces on ech body into components Use convenient choice of x,y xes Use the rules for finding vector components (Ch. 3) So Wht is the velocity of the sled w/ mx (from rest) fter 5 minutes? F N T y f MAX m f Xyou x Fg Box weighs 10kg x 9.8m/s 2 = 90N y: N + T sin -mg = m Y = 0 N = mg - T sin 30 = 60 N Fy NET = N - mg Fy NET = 60N - 98N 1. The box s net force in y xis is -38N nd it does not lift No ccelertion=no movement y T = 40 N x N T m = 30 mg PUSHIN ON THE BOX This imge cnnot currently be displyed. Mx pulls with = 30N x: T cos = m X 40N cos 30 = 34.6 N Fx NET = You - Mx Fx NET = 34.6N - 30N 2. The box s net force in x xis is +4.6N ccelertion = movement m N mg y T = 40 N x T Elevtor Problem Hve students determine tht the force e of grvity is 9.8N down nd the t if the scle is set to 9.8 tht the force norml up is 19.6N up nd the Net is 9.8 N up on 1Kg, so the net ccelertion is 9.8 m/s2. Also show tht s the brick drops tht the force norml would go to zero nd so the reding would go to zero. Apply Fx NET = m 4.6N = (10kg+25kg).13m/s 2 = 10

TENSION Mke sure your clcultors re not in rdins mode Mke sure tht you execute trig functions ppropritely Tension Forces Creted by pplying force to rope, cble Equl t ll points in rope Wht does net force do? Is T 1 =T 2 =T 3? http://www.youtube.com/w tch?feture=plyer_detilp ge&v=gf3iz7b95-8 Tension nd Angles Drw FBD: T 1 sin T 1 cos T 1 T 2 T 2 sin T 2 cos mg Since the box isn t going nywhere, F x,net = 0 nd F y,net = 0 F x,net = T 1 cos - T 2 cos = 0 T 1 = T 2 F y,net = T 1 sin + T 2 sin -F g = 0 mg T 1 = T 2 = 2 sin i j A 300N box is suspended from the ceiling by two ropes mking n ngle with the horizontl. If thet is 50 o wht is the tension on ech wire? 300N 11

This imge cnnot currently be displyed. If thet is 50 o wht is the tension on ech wire? X AXIS A 125 N trffic light is supported s shown. Drw FBD for the hoop nd the light. Find tension in ech cble. FBD y AXIS T 1 T 2 300N Mr. Cistowlkcross high wire strung horizontlly between the SHS towers 10.0 m prt. The sg in the rope when he is t the midpoint is 10.0º s shown. If hismssis50.0kg,whtisthetensionintheropet this point? FBD T 3 must equl the weight of the light, 125 N Find x nd y components of T 1 nd T 2 : The y components must equl 125 N The x components must equl zero. 1 2 10o 10 o Two equtions, two unknowns. Solve for T 1 in first: T 1 = 0.7002 T 2 Plug in this vlue into second eqution: If thet is 10 o wht is the tension on ech wire? X AXIS y AXIS T 2 = 102 N T1 = 0.7002 T2 T 1 = 71.4 N 12

If the Fg is 125 N, Does this men tht the T on rope is twice tht. FBD T 1 T 2 70 20 F g Pull on Student demo X AXIS http://www.youtube.com/wtch?v=x3j8zmhghz8&fet ure=relted&sfety_mode=true&persist_sfety_mode= 1&sfe=ctive y AXIS=Fr X AXIS Two helicopters lift HUMVEE of mss 11440kg. Determine T 2 nd T 2 http://www.youtube.com/wtch?v=lrkn8dcyyre&feture=relted 70 20 T1 T2 13

Pulleys IN YOUR NOTES ANSWER THIS QUESTION: WHAT WILL HAPPEN IF WE LET THESE MASSES HANG HERE OVER TIME? WHAT HAPPENS IF WE MESS UP? PULLEYS Grndm cuses tension Tke rope nd hve FOUR students pull it. Who is pulling hrder. It is not moving, so wht do we know bout forces. FORCES ARE BALANCED. Tke the rope nd hve two students pull down on it. If I tell two students to climb the rope, nd they don t weigh the sme mount wht will hppen (T&T) Hve students strt by conducting tug of wr hve nother stnd perpendiculr to the rope nd keep the rope from, thn wrpping it round the meter stick. Then hve him hold rolling pin. 14

Tension Motion Cse 1: COMPLEX SLED A tow truck hs to pull motor home nd triler (13000Kg nd 1100Kg) cross n frictionless rod t 1000N. Wht is, T1, nd T2? m 2 m 1 T 2 T 1 1100Kg 13000 Kg Does the tension on the rope increse s the semi strts to move? http://www.youtube.com/wtch?nr=1&v=vfkt6qhbn8&feture=endscreen Why does the fther wnt to be cut free of the rope? Does ech rope feel different Tension? If the reindeer pulls n elf sleigh nd Snt sleigh (20Kg nd 100Kg) t 500N wht is? Wht is, T1, nd T2? 15

The imge prt with reltionship ID rid5 ws not found in the file. 1 2 2 blocks hng in n elevtor s shown. The elevtor ccelertes upwrd t 3.00 m/s 2. Find the tension in ech rope. Becuse we hve & g Find Tension 2: FBD How much tension must rope withstnd if it is used to ccelerte Crider s Ride ( 1200-kg cr) verticlly upwrd t 0.80m/s 2? FBD T 2 = 256 N 12720N FBD HERE COMES THE PULLEY Find Tension 1: NOW WE ARE JUST GOING TO TILT THE PREVIOUS PROBLEMS TO THE X AXIS AND LET THE SYSTEM WORK WITH / AGAINST GRAVITY. T 1 = 512 N A 12.0-kg bucket is rised verticlly by rope in which there is 163 N of tension t given instnt. Wht is the ccelertion of the bucket? Is it up or down? FBD The cble supporting 2125-kg elevtor hs mximum strength of 21,750 N. Wht mximum upwrd ccelertion cn it give the elevtor without breking? FBD 16

Tension Motion Cse 2: PULLEY Msses m 1 nd m 2 re ttched to n idel mssless string nd hung s shown round n idel mssless pulley. T 1 How to set ny pulley style problems up Light: mss is ccelerting wy from grvity the Tension on the Rope increses hevy mss, M 2, will fll, lifting the smller mss, m 1. velocity of mss m 2 is equl in mgnitude to the velocity of mss m 1, but opposite in direction. ccelertion of mss m 1 is equl in mgnitude to the ccelertion of mss m 2, but opposite in direction. 1 T 1 T 2 m 1 m 2 2 1 m 1 g m 1 mss is ccelerting towrds grvity pulley cuses Tension on the Rope to decrese T 2 m 2 2 m 2 g Thrust 1 Thrust 1 PULLEY 1 m 1 2 m 1 g m 2 g T T m 1 m 2 And plug into either formuls for T The downwrd will reduce the tension on the hevy mss The upwrd will increse the tension on the light mss A pulley llows the T to be equl by opposite Light T 1 T 2 1 m 1 m 2 2 m 1 g 1 = -2 m 2 g Hevy Two msses, 4.00 kg nd 5.25 kg re connected by light string to frictionless pulley s shown. 1) Drw the FBD. 2) Find the ccelertion on the system the tension in the string. 2) Find Tension on the string. Remember tht tension is the sme for both. FBD 17

FIND The forces on the rising mss, use up s positive: CONNECTION For the flling mss, down is positive Add the 2 equtions: Grph the elevtors motion FIND T A pulley 6.0 m high nd two msses (3.0 nd 5.0 Kg) nd relesed from ech t 2.0m. Wht will be the mx height of the light mss fter the hevy strikes the ground T T 44.5 N is the vlue for both tensions IF WE KNOW THE WE NOW GO BACK TO THE MOTION FORMULAS 3 msses hng s shown, they re connected by light strings nd your bsic frictionless pulley. find the ccelertion of ech mss nd tensions in the 2 strings. DRAW THE FREE BODY DIAG The mgnitude of ccelertion for ech mss is = Flling msses (left side) down is +. Right side up is + F = m F m 1 : m 1 g T 1 = m 1 F m 2 : m 2 g + T 1 T 2 = m 2 F m 3 : T 2 m 3 g = m 3 18

Add the equtions: m 1 g + m 2 g m 3 g = m 1 + m 2 + m 3 = 1.63m/s 2 Find Tensions: T2 T 2 A pulley with weights of 89.94 g nd 90.31g re suspended from pulley tht is 1.71m from the floor. A key is ttched to the hevier mss. The weights re held evenly 77.5cm from the floor. When relesed it tkes the hevier mss 2.20s to hit the floor. )Wht is the weight of the key. b)wht is the zenith of the lighter mss? T 3 AP Question Demo: Mss of Key Find the net ccelertion Find the Mss of the Key Find the Tension on the cble T 1 1 m 1 m 1 g Tension Motion Cse 3: Overhng Left: no net force in y since Fn=Fg Right mss is ccelerting towrds grvity pulley cuses Tension on the Rope to decrese T 2 m 2 2 Rerrnge to find ccelertion m 2 g 19

Tension Motion Cse 3: Overhng LETS CHECK THIS If 1 = 0 nd 2 = 90, A 20.0 kg crt with very low friction wheels sits on tble. A light string is ttched to it nd runs over low friction pulley to 0.0150 kg wter bottle. A) Drw FBD b) Wht is the ccelertion experienced by the crt? C) Wht is the fll time if tble is 6m high (ssume rope limit.) Wht will be the velocity the crt lunches off the tble? Forces on ech object: Crt: F x = T = m 1 Attched bodies on two inclined plnes smooth peg m 1 m 2 1 2 All surfces frictionless Solve for ccelertion: How will the bodies move? = 0.00734 m/s 2 Solve for t fll: From the free body digrms for ech body, nd the chosen coordinte system for ech block, we cn pply Newton s Second Lw: Tking x components: x y x y N T 1 T 2 1) T 1 -m 1 g sin 1 = m 1 1X 2) T 2 -m 2 g sin 2 = m 2 2X N m 2 m 1 But T 1 = T 2 = T nd - 1X = 2X = (constrints) m 1 g 1 2 m 2 g 20

The imge prt with reltionship ID rid24 ws not found in the file. Specil Cse 4: m 1 m 2 m sin m sin 2 2 1 1 g m m 1 2 g 1 1 1 2 g t=-0.41s nd 0.58s Tigger s Demise m 2 m 1 2 1 Fp Ff Qudrtic Eqution Newton s Third Lw: Action / Rection Whenever one object exerts force on second object, the second object exerts n equl nd opposite force on the first object. A rection force does not cncel n ction force. Although the ction nd rection forces re equl In strength, they re opposite in direction, giving net force of zero on the object itself (not the contct source) Cn you touch without being touched? Cn one hnd clp? 21

Action-Rection Pirs: On Different Bodies F A,B = - F B,A. is true for contct forces s well: F m,w F w,m F f,m F m,f Action-Rection Pirs: On Different Bodies Since F m,b = -F b,m, why isn t F net = 0 nd = 0??? F m,b F b,m ice Consider only the box s the system! As long s the box retins its shpe (doesn t crush) F b,m = m box = F box Since F m,b = -F b,m, why isn t F net = 0 nd = 0? F b,m F m,b?? F m,b F b,m box F m,b F b,m ice ice 22

Inclines F n F g Is the norml force F N equl & opposite to the weight F G? NO!!!!!!!!!!!! WHICH AXIS ARE THESE FORCES We tilt the x,y xis THEY ARE AGAINST EACH OTHER, DOES IT MATTER? WHY IS IT HARDER FOR HIM TO PUSH AS TIME PASSES? Wht is the mechnicl dvntge of the rmp or incline plne? Inclines: Which wy is up? F N force weight Wht do you notice bout the difference between these two rmps? Fg Fg For n object on slope, resolve its weight into component tht cuses ccelertion long the slope nd component tht presses it ginst the slope. Must resolve weight correctly before summing forces in x, y 23

Problem: Inclined plne A block of mss m slides down frictionless rmp tht mkes ngle θ with respect to the horizontl. Wht is its ccelertion? Rmp elevted t 28.0 ngle nd is kept from sliding down by rope tied to secure block s shown If block hs weight of 225 N, 1. Drw the FBD 2. Find Tension on rope holding it up? m θ Inclined plne... Define convenient xes prllel nd perpendiculr to plne: Accelertion is in x direction only. m y x F x = 0 nd F y = 0 We cn solve T by isolting X Component In x direction: T is blnced by force down the rmp F x = T F g sin = 0 T = 225 N sin 28.0 T = 106 N Inclined plne... Consider x nd y components seprtely: x:mg sin =m = g sin Rmp elevted t 52.0 ngle If the block hs weight of 50 N, 1. Find Norml Force 2. Find Prllel Force? y:n - mg cos = 0 mg sin N = mg cos m N mg cos mg 24

1) Norml Force An snowbll is rolled down snowbnk of 30 o It is 300 m bove the mountin bse if Mr C is stnding t the bottom how much time does he hve to get out of the wy? 2) Prllel Force 30 30 A 5.00 kg bll rolls down 18.0 rmp. Wht is the ccelertion of the bll? Ignore friction. If the rmp is 2.00 m long, how much time to rech the bottom? A bowling bll is pushed down frictionless hill of 50 o. It is pushed with n initil velocity of 5m/s how fst will it hit the pin tht is 16m wy? 30 50 t = 1.15 s of the object on the slope (in the bsence of friction) does not depend on the object's mss. 25

Solve for t slide: LETS CHECK THIS 1. 2. 3. 4. A= t= 5. = Vf b) c) Surfce Friction... Friction is cused by the microscopic interctions between the two surfces: Friction forces ALWAYS oppose ctul or ttempted motion between 2 surfces http://www.youtube.com/wtch?v=48zfd_zwy8&sfety_mode=true&persist_sfety_ mode=1&sfe=ctive Friction forces creted by microscopic cold-welding between surfces Friction Friction Wht does it do? It opposes reltive motion! How do we chrcterize this in terms we hve lerned? Friction results in force in the direction opposite to the direction of reltive motion! N j F APPLIED m i f FRICTION mg 26

Forces of Friction: f Dynmics: F f = * F N When n object is in motion on surfce or through viscous medium, there will be resistnce to the motion. This resistnce is clled the force of friction This is due to the interctions between the object nd its environment Force of sttic friction: f s Force of kinetic friction: f k Direction: long the surfce, opposite the direction of the intended motion in direction opposite velocity if moving in direction vector sum of other forces if sttionry y : F N = mg x : F net = F F f = m F F N = m so F mg = m F N m y x F f mg The coefficients of friction re nerly independent of the re of contct Two types of friction: Kinetic friction: sliding motion f k = k F N Sttic friction: no motion f s s F N Surfce Friction... FBD Force of friction cts to oppose reltive motion: Prllel to surfce. Perpendiculr to Norml force. N y F m x f F mg WE NEED TO DETERMINE THE NET FORCE 27

Sttic & Kinetic Friction A box of mss m =1500 kg is t rest on floor. The coefficient of kinetic friction between the floor nd the box is M k = 0.331. A rope is ttched to the box nd pulled t n ngle of = 20 o bove horizontl with tension T = 600 N. Find the. T sttic friction ( s = 0.3 ) m N T f FR m Sttic frictionl forces lwys greter thn sliding ones The box will move if T cos -f FR > 0 mg A box of mss m =10.21 kg is t rest on floor. The coefficient of sttic friction between the floor nd the box is m s = 0.4. A rope is ttched to the box nd pulled t n ngle of = 30 o bove horizontl with tension T T = 40 N. Drw FBD. Does the box move? With wht? sttic friction ( s = 0.4 ) m N T f FR m The box will move if T cos -f FR > 0 mg The box will move if T cos -f FR > 0 So T cos > f f nd the box does move A 25.0 N wood block is pulled cross wooden tble t constnt speed. Wht is the force needed to do this? Use 0.35 s k. The box will move if T cos -f FR > 0 The net force must be zero in the x F x = f k F = 0 so F = f k Fy = n F g = 0 n = F g nd y direction So T cos > f f nd the box does not move F f only equls the 563 Force of tension f k = 8.8 N 28

A 10.0 kg block is plced on 39.1 o inclined plne with coeff of friction of 0.25. ) Drw free-body digrm. b) Lbel F W, F P, nd F N.. c) Clculte F P d) Clculte F N nd F f Into the incline ) Up the incline 39.1 o If there is no incline (0deg) then the cos is mx or =1 Wht is the force due to friction on 30.7 kg box sitting on 48.0 o incline where the coefficient of friction is 0.311?... down the incline 29

A 10.0 kg box is on 22.0 o incline where the coefficient of sliding friction is 0.782. Find the ) FBD b) norml force & prllel force, c) force due to kinetic friction, d) net force, e) ccelertion, nd f) the time when the box will be 30.0 m down the incline, ll ssuming zero velocity t the strt. A box is on 55.0 o incline where the coefficient of sliding friction is 0.6. Find the ccelertion, nd the finl velocity fter 2 minute ssuming zero velocity t the strt. You cn solve this problem without knowing the mss To Find the Norml Force A 50kg child plummets down wter slide tht is 200 meters long. The slide hs n verge slope of 25 degrees, with pool t the bottom. The coeff of kinetic friction of slide is.1. The child slides 25 meters on the wter before stopping (then sinking). Wht ws the coefficient of friction on the wter? To Find the Friction Force To Find the Prllel Force Drw Both FBDs To Find the Net Force No Accelertion 30

A 50kg child plummets down wter slide tht is 200 meters long. The slide hs n verge slope of 25 degrees, with pool t the bottom. The coeff of kinetic friction of slide is.1. The child slides 25 meters on the wter before stopping (then sinking). Wht ws the coefficient of friction on the wter? A 150kg child plummets down wter slide tht is 200 meters long. The slide hs n verge slope of 25 degrees, with pool t the bottom. The coeff of kinetic friction of slide is.4. The child slides 25 meters on the wter before stopping (then sinking). Wht ws the coefficient of friction on the wter? Drw Both FBDs Drw Both FBDs 31

Size of ski does not ffect frictionl force Terminl Velocity =0m/s 2 Reched when F drg = F grv (= mg) For 75 kg person 0.5 m 2 v term 50 m/s, or 110 m.p.h. (reched in bout 5 seconds, in 125 m) ctully tkes slightly longer, becuse ccelertion is reduced from the nominl 9.8 m/s 2 s you begin to encounter drg Free fll only lsts few seconds, even for skydivers http://www.youtube.com/wtch?v=ayz_k3mwq6a&feture=grec_index&sfety_mode=true&persist_sfety_mode=1&sfe=c tive http://www.youtube.com/wtch?v=fhtvda0w34i&feture=g-logo-xit 32

http://www.youtube.com/wtch?v=ur40o6nqhsw&sfety _mode=true&persist_sfety_mode=1&sfe=ctive EXAM TIME Just strts to move t 20deg At 25 deg it moves t constnt ccelertion down incline in 5 sec FORMULAS GIVE ON AP EXAM Find the s nd k Find the s =.34 nd k=.36 Just strts to move t 20deg At 25 deg it moves t constnt ccelertion down incline in 5 sec 33