UEquations of MotionU
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- Lenard Strickland
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1 UChapter Mtin in One Dimensin When a particle mves, its displacement is defined by its change in psitin. As it mves frm an initial psitin xi t a final psitin x f, its displacement is given by x = x x f The average velcity f a particle is defined as the particle s displacement x divided by the time interval t during which that displacement ccurred: x ν x = t The average velcity f a particle mving in ne dimensin can be psitive r negative, depending n the sign f the displacement. The velcity and speed are different. Cnsider a marathn runner wh runs 40 km in hurs, ends up at his starting pint. His average velcity is zer. The average speed is a scalar quantity and defined as the ttal distance traveled divided by it takes = 40 km / h = 0 km/h i The instantaneus velcity vx equals the limiting value f the rati x/ t as t appraches zer. x dx ν x = lim = x 0 t dt The average acceleratin f the particle is defined as the change in velcity vx divided by the time interval t during which that change ccurred: ν a x = t dν x The instantaneus acceleratin f the particle is defined as a = dt UEquatins f MtinU معادلات الحركة في خط مستقيم وبتسارع ثابت كل معادلة يمكن ا ستخدامها لا يجاد مجهول واحد فقط
2 υ= υ + at υ (1) 1 x x = υt + at () υ = + ax ( x ) (3) 1 x x = ( υ + υ ) t (4) 1 x x = υt at (5) معادلات السقوط الحر : EquatinsU U- Free Fall هذه المعادلات تعتبر معادلات حركة في الا تجاة الرا سي (y) وا ستبدال قيمة التسارع (g) بقيمة تسارع الجاذبية υ= υ + at υ (1) 1 y y = υt + gt () υ = + g ( y y ) (3) 1 y y = ( υ + υ ) t (4) 1 y y = υt at (5) y = 0, a = - g, y is negative y = 0, a = + g, y is psitive UPrblems في في حالة السقوط الحر حالة الصعود ا لى ا على 1. A car, initially at rest, travels 0 m in 4 s alng a straight line with cnstant acceleratin. The acceleratin f the car is: A. 0.4 m/s B. 1.3 m/s C..5 m/s D. 4.9 m/s E. 9.8 m/s ans: C
3 . A racing car traveling with cnstant acceleratin increases its speed frm 10 m/s t 50 m/s ver a distance f 60 m. Hw lng des this take? A..0 s B. 4.0 s C. 5.0 s D. 8.0 s E. The time cannt be calculated since the speed is nt cnstant ans: B 3. A car starts frm rest and ges dwn a slpe with a cnstant acceleratin f 5 m/s. After 5 s the car reaches the bttm f the hill. Its speed at the bttm f the hill, in meters per secnd, is: A. 1 B. 1.5 C. 5 D. 50 E. 160 ans: C 4. A car mving with an initial velcity f 5 m/s nrth has a cnstant acceleratin f 3 m/s suth. After 6 secnds its velcity will be: A. 7 m/s nrth B. 7 m/s suth C. 43 m/s nrth D. 0 m/s nrth E. 0 m/s suth ans: A 5. Hw far des a car travel in 6 s if its initial velcity is m/s and its acceleratin is m/s in the frward directin? A. 1 m B. 14 m C. 4 m D. 36 m E. 48 m ans: E 6. A baseball is thrwn vertically int the air. The acceleratin f the ball at its highest pint is:
4 A. zer B. g, dwn C. g, up D. g, dwn E. g, up ans: B 7. An bject is thrwn straight up frm grund level with a speed f 50 m/s. If g = 10 m/s its distance abve grund level after 6.0 s is: A m B. 70 m C. 330 m D. 480 m E. nne f these ans: E 8. At a lcatin where g = 9.80 m/s, an bject is thrwn vertically dwn with an initial speed f 1 m/s. After 5 s the bject will travele A. 15 m B m C. 45 m D. 50 m E. 55 m ans: B 9. An bject is thrwn vertically upward at 35 m/s. Taking g = 10 m/s, the velcity f the bject after 5 s is: A. 7.0 m/s up B. 15 m/s dwn C. 15 m/s up D. 85 m/s dwn E. 85 m/s up ans: B 10. A heavy ball falls freely, starting frm rest. Between the third and furth secnd f time it travels a distance f: A. 4.9 m B. 9.8 m C. 9.4 m D m
5 E m ans: D 11. An bject drpped frm the windw f a tall building hits the grund in 1 s. If its acceleratin is 9.80 m/s, the height f the windw abve the grund is: A. 9.4 m B m C. 118 m D. 353 m E. 706 m ans: E 1. A baseball is hit straight up and is caught by the catcher.0 s later. The maximum height f the ball during this interval is: A. 4.9 m B. 7.4 m C. 9.8 m D. 1.6 m E m ans: A 13. An bject is thrwn straight dwn with an initial speed f 4 m/s frm a windw which is 8 m abve the grund. The time it takes the bject t reach the grund is: A s B s C. 1.3 s D. 1.7 s E..0 s ans: B 14. A stne is released frm rest frm the edge f a building rf 190 m abve the grund. The speed f the stne, just befre striking the grund, is: A. 43 m/s B. 61 m/s C. 10 m/s D. 190 m/s E m/s
6 ans: B Prblem : A particle mves alng the x axis accrding t the equatin x = + 3t t where x is in meters and t is in secnds. At t = 3 s, find (a) the psitin f the particle, (b) its velcity, and (c) its acceleratin.
7 الا حداثيات القطبية Plar Cridnates Starting with the plar crdinates f any pint, we can btain the cartesian crdinates, using the equatins x = r cs θ y = r sin θ y tanθ = x r = x + y # Prblem The cartesian crdinates f a pint in the x-y plane is (- 3.5, -.5) m. Find the plar crdinates f this pint. USlutin : r = x + y = ( 3.5) + (.5) = 4.3 m y.5 tanθ = = = x 3.5 θ = 16 UVectr and Scaler Quantities A scalar quantity is specified by a single value with an apprpriate unit and has n directin. Examples f scalar quantities are vlume, mass, and time intervals. A vectr quantity has bth magnitude and directin, such as displacement, velcity and frce.
8 1- UEquality SOME PROPERTIES OF VECTORS These fur vectrs are equal because they have equal lengths and pint in the same - UAdding Vectrs When vectr B is added t vectr A, the resultant R is the vectr that runs frm the tail f A t the tip f B. المحصلة هي المتجة الذي يبدا من ذيل المتجة A وينتهي عند را س المتجة B 3- Cmmutative law A + B = B + A
9 A + ( B + C) = ( A + B) + C طرح المتجهات 4- USubtracting VectrsU The peratin f vectr subtractin makes use f the negative vectr. The peratin A B = A + ( B) as vectr B added t A. Prblems P1- The vectr A is: A. greater than A in magnitude B. less than A in magnitude C. in the same directin as A D. in the directin ppsite t A E. perpendicular t A P- A vectr has a magnitude f 1. When its tail is at the rigin it lies between the psitive x axis and the negative y axis and makes an angle f 30 with the x axis. Its y cmpnent is: A. 6/ 3 B. 6 3 C. 6 D. 6 E. 1
10 3- Vectr V 3 in the diagram is equal t AV. 1 V B. V1 + V C. V V1 D. V1csθ E. V / csθ 1 5- Multiplying a Vectr by a Scalar If vectr A is multiplied by a psitive scalar quantity m, then the prduct m A is a vectr that has the same directin as A and magnitude m A. 6- Cmpnents f a Vectr and Unit Vectrs : If a vectr A has x-cmpnent A x = A cs θ y-cmpnent A y = A sin θ the vectr can be expressed in unit vectr frm as A = A x i + A y j In this ntatin, i unit vectr pinting psitive x- directin, j unit vectr pinting psitive y-directin. i = j = 1
11 Mtin in Tw Dimensin The displacement f a particle as it mves frm A t B in the time interval t = tf ti is equal t the vectr r = r r f i The average velcity f a particle during time interval t as the displacement f the particle r divided by that time interval r v = average velcity is independent f path t السرعة اللحظية The instantaneus velcity ν is defined as r dr ν = lim = (1) t 0 t dt التسارع اللحظي The instantaneus acceleratin a is defined as dν d ( dr / dt ) d r a = = = () dt dt dt Tw-Dimensinal Mtin with Cnstant Acceleratin Let us cnsider tw-dimensinal mtin during which the acceleratin remains cnstant in bth magnitude and directin. The psitin vectr f a particle mving in the xy plane is r= xi. ˆ + yj. ˆ (3) the velcity f the particle is given by ν = ν iˆ+ ν jˆ x y (4)
12 If a particle mves with cnstant acceleratin a and has velcity v i and psitin r i at t = 0, its velcity and psitin vectrs at sme later time t are νf = νi + at. 1 r f = r i + νi. t + at. Prlebms: 1. Which f the fllwing is a scalar quantity? A. Speed B. Velcity C. Displacement D. Acceleratin E. Nne f these. Which f the fllwing is a vectr quantity? A. Mass B. Density C. Speed D. Temperature E. Nne f these 3. A particle ges frm (x = m, y = 3m) t (x = 3m, y = 1m), Its displacement is: A. (1 m)ˆi + ( m)ˆj B. (5 m)ˆi (4 m)ˆj C. (5 m)ˆi + (4 m)ˆj D. (1 m)ˆi ( m)ˆj E. (5 m)ˆi ( m)ˆj المقذوفات Prjectiles (5) (6)
13 The prjectile leaves the rigin with a velcity v i The velcity v changes with time in bth magnitude and directin. The x cmpnent f velcity remains cnstant with time The y cmpnent f velcity is zer at the peak f the path معادلة المدى ا فقي معادلة ا قصى ا رتفاع h ν = i sin g θ R νi sin θ = g A prjectile fired frm the rigin with initial speed f 50 m/s with different θ The maximum range ccurs when θ = 45 Prblems 1- A large cannn is fired frm grund level at an angle f 30 abve the hrizntal. The muzzle speed is 980 m/s. The hrizntal distance the prjectile will travel befre striking the grund, assume g = 9.8 m/s A. 4.3 km B. 8.5 km C. 43 km D. 85 km E. 170 km ans: D - A by n a rugh f building 0 m high kicks a ball hrizntally utward with a speed f 0 m/s. It strikes the grund at hrizntal distance frm the wall, assume g = 10 m/s A. 10 m B. 40 m C. 50 m D m E. nne f these
14 ans: B 3- A prjectile is fired frm grund level with an initial velcity that has a vertical cmpnent f 0 m/s and a hrizntal cmpnent f 30 m/s. Using g = 10 m/s, the distance frm launching t landing pints is: A. 40 m B. 60 m C. 80 m D. 10 m E. 180 m Unifrm Circular Mtin A particle is in unifrm circular mtin if it travels arund a circle at cnstant speed. Althugh the speed des nt change, the particle is accelerated because the velcity changes directin. The acceleratin is directed tward center f the circle ( الداي رة (مركز and given by ν (تسارع مركزي ( acceleratin a r = centripetal r Where r is the radius f the circle and ν is the speed f the particle. The time required t cmplete ne circle is called π r The peridic time ( الدوري (الزمن T = v 1 The frequency ( الدورات في الثانية الواحدة (عدد, f = T Prblems 1- A car runds a 0 m radius curve at 10 m/s. The magnitude f its acceleratin is: A. 0 B. 0. m/s C. 5 m/s D. 40 m/s E. 400 m/s ans: C - A stne is tied t 50 cm string ( (خيط and whirled ( ي( دور at cnstant speed f 4 m/s in a vertical circle. Its acceleratin at the bttm f the circle is A. 9.8 m/s, up
15 B. 9.8 m/s, dwn C. 8.0 m/s, up D. 3 m/s, up E. 3 m/s, dwn ans: D 3- A stne is tied t the end f a string and is swung with cnstant speed arund a hrizntal circle with a radius f 1.5 m. If it makes tw cmplete revlutins each secnd, its acceleratin is A. 0.4 m/s B..4 m/s C. 4 m/s D. 40 m/s E. 400 m/s ans: D Relative Velcity and Relative Acceleratin The psitin f a particle A relative t the S frame with the psitin vectr r and that relative t the S frame with the psitin vectr ' r. The vectrs r and ' r are related t each ther at time t r = r v t dr dr = v dt dt v = v v
16 Example: A bat heading nrth crsses a river with speed f 10 km/h relative t the water. The river has speed f 5.00 km/h t east relative t the Earth. Determine velcity f the bat relative t an bserver standing n bank The relatinship between the three velcities is vbe = vbr + v re be br re v = v + v = (10) + (5) = 15 v re tanθ = =, θ = tan ( ) v 10 br
17 1- Newtn s first law : Laws f Mtin In the absence f external frces, an bject at rest remains at rest and an bject in mtin cntinues in mtin with the same velcity. - Newtn s secnd law : The acceleratin f an bject is directly prprtinal t the net frce acting n it and inversely prprtinal t its mass. F = ma The SI unit f frce is the newtn, which is defined as the frce acting n a 1-kg mass, prduces an acceleratin f 1 m/s. Example : 1N = 1 kg. m / s A hckey puck having a mass f 0.30 kg slides n a hrizntal, frictinless surface f an ice rink. Tw frces act n the puck, as shwn in Figure. Determine bth the magnitude and directin f the puck s acceleratin. 1cs 0 Σ F = F + F cs60 = x = 5 cs0 + 8 cs60 = 8.7 N 1 Σ F = F sin 0 + F sin 60 = y = 5 sin sin 60 = 5. N ΣFx 8.7 N ax = = = 9 m / s m 0.3kg ΣFy 5. N ay = = = 17 m / s m 0.3kg a = (9) + (17) = 34 m / s 1 ay 1 17 θ = tan ( ) = tan ( ) = 30 a 9 x
18 1. A frce f 1N is: A. 1 kg/s B. 1 kg m/s C. 1 kg m/s D. 1 kg m /s E. 1 kg m /s ans: C Prblems. When a certain frce is applied t the standard kilgram its acceleratin is 5 m/s. When the same frce is applied t anther bject its acceleratin is ne-fifth as much. The mass f the bject is: A. 0. kg B. 0.5 kg C. 1 kg D. 5 kg E. 10 kg ans: D 3. A cnstant frce f 8 N is exerted ( (تو ثر fr 4 s n a 16 kg bject initially at rest. The change in speed f this bject will be: A. 0.5 m/s B. m/s C. 4 m/s D. 8 m/s E. 3 m/s ans: B 4. Tw frces, given by F 1 = (- 6i - 4j) N and F = (- 3i + 7j) N, act n a particle f mass kg that is initially at rest at the rigin. (a) What are the cmpnents f the particle s velcity at t = 10 s. (b) In what directin is the particle mving at t = 10 s, (c) What displacement des the particle make. The weight f an bject is defined as Frce f Gravity and Weight Fg = mg where g is the gravity acceleratin equal = 9.8 m/s r 10 m/s Because weight Fg = mg, we can cmpare the masses f tw bjects by measuring their weights n a spring scale. The rati f the weights f tw bjects equals the rati f their masses.
19 Newtn s Third Law : If tw bjects interact, the frce F 1 exerted by bject 1 n bject is equal in magnitude and ppsite in directin t the frce F 1 exerted by bject n bject 1: F = The frce F hn exerted by the hammer n the nail is equal and ppsite the frce F nh exerted by the nail n the hammer. F 1 1 Weighings in Elevatr A persn weighs a fish f mass m n a spring scale attached t the elevatr. If the elevatr accelerates either upward r dwnward, the scale gives the weight f the fish. Slutin 1- If the elevatr is either at rest r mving at cnstant velcity, Newtn s secnd law gives Fg = mg= T - If the elevatr mves upward with an acceleratin, Newtn s secnd law applied t the fish gives the net frce
20 Fy = T mg = m ay T = mg + m ay = m ( g + ay ) () The scale reading T is greater than the weight mg if a y is upward (+), and the reading is less than mg if a y is dwnward ( ). Prblems : kg man stands in an elevatr that has a dwnward acceleratin f 1.4 m/s. The frce exerted by him n the flr is abut: A. zer B. 90N C. 760N D. 880N E. 1010N ans: C Atwd s Machine When tw bjects f unequal mass are hung ( (تعلق vertically ver a frictinless pulley, the arrangement is called an Atwd machine.
21 When Newtn s secnd law is applied t bject 1, we btain T mg= ma (1) 1 1 y Similarly, fr bject we find m g T = m a () y When () added t (1), T is drpped and get m m1 ay = ( ) g m + m 1 When (3) is substituted int (1), we btain mm 1 T = ( ) g m + m 1 (3) (4). 70 N blck and 35 N blck are cnnected by a string. If the pulley is massless and the surface is frictinless, the magnitude f the acceleratin f the 35 N blck is:
22 A. 1.6 m/s B. 3.3 m/s C. 4.9 m/s D. 6.7 m/s E. 9.8 m/s ans: B 3. Tw blcks (A and B) are in cntact n a hrizntal frictinless surface. A 36 N cnstant frce is applied t A. The frce f A n B is A. 1.5 N B. 6.0 N C. 9 N D. 30 N E. 36 N Ans: D a x = m F + m 1 4. m 1 = kg, m = 6 kg θ = 55 frictinless pulley Find : (a) the acceleratins f the masses, (b) the tensin in the string, (c) the speed f each mass.00 s after being released frm rest. Fr the first bdy T mg= ma (1) 1 1 Fr the secnd bdy m gsinθ T = m a () Add equatins (1) and (), gives the acceleratin. Substitute the value f a in the first equatin gives the tensin in the string. T find the speed after s apply the law, v = v + at = 0+ a = m / s
23 Frce f Frictin T push a heavy desk n rugh flr, it takes a greater frce t start mving than it takes keep mving. Increasing F, the magnitude f F s increases alng with it, keeping the bk in place. When it is n the verge f mving, F s has a maximum value. When F exceeds F sm, the bk accelerates t the right. Once the bk is in mtin, the retarding frictinal frce is called kinetic frictin F k and smaller than F sm. right. If F = F k, the bk mves t the right with cnstant speed. If F > F k, the unbalanced frce (F - F k ) accelerates the desk t the If the applied frce F is remved, then the frictinal frce F k decelerating ( السرعة (تقلل the desk and eventually brings it t rest. Fs = µ s FN, where µ s is the cefficient f static frictin Fk = µ k FN, where µ k is the cefficient f kinetic frictin F is the nrmal frce N Prblems 1. A blck mves with cnstant velcity n a hrizntal rugh surface. The frictinal frce necessary t keep a cnstant velcity is:
24 A. 0 B. N, leftward C. N, rightward D. slightly mre than N, leftward E. slightly less than N, leftward ans: B. A persn pushes hrizntally with a frce f 0 N n a 55 kg crate t mve it acrss a level flr. The cefficient f kinetic frictin is 0.35 what is the magnitude f the frictinal frce and the acceleratin kg crate is n an incline that makes an angle f 30 with the hrizntal. If the cefficient f static frictin is 0.5, the maximum frce that can be applied parallel t the plane withut mving the crate is: A. 0 B. 3.3 N C. 30 N D. 46 N E. 55 N ans: D 4. The system shwn remains at rest. Each blck weighs 0 N. The frce f frictin n the upper blck is: A. 4N B. 8N C. 1N D. 16N E. 0N ans: B
25 5. 9 kg hanging weight is cnnected by a string ver a pulley t a 5 kg blck that is sliding n a flat table. If the cefficient f kinetic frictin is 0., find the tensin in the string. 6. Tw blcks cnnected by a rpe f negligible mass are being dragged by a hrizntal frce F = 68 N, m 1 = 1 kg, m = 18 kg, and the cefficient f kinetic frictin between each blck and the surface is 0.1 Determine : (a) the tensin T, (b) the acceleratin f the system. 6. The magnitude f the frce required t cause a 0.04 kg bject t mve at 0.6 m/s in a circle f radius 1.0 m is: A N B N C. 1.4π 10 N D..4π 10 N E. 3.13N ans: B kg stne is attached t a string and swung in a circle f radius 0.6m n a hrizntal and frictinless surface. If the stne makes 150 revlutins per minute, the tensin frce f the string n the stne is: A N B. 0. N
26 C. 0.9 N D N E. 30 N ans: E
27 Wrk and Kinetic Energy Kinetic Energy : Fr an bject f mass m and speed ν, its kinetic energy K is given by K 1 mν = J (Jule = 1 kg. m /s ) Wrk dne : Wrk dne by a cnstant frce is the prduct f the cmpnent f the frce in the directin f displacement and the magnitude f the displacement: W = F. d csθ J If an applied frce F acts alng the directin f the displacement, then θ = 0, and cs θ = 1. The abve equatin gives W = F. d Wrk is scalar quantity, and its units are frce multiplied by length. (N.m). The change in kinetic energy = net wrk dne n the particle K = W K K = W f i Example : A 0.3 kg ball has a speed f 15.0 m/s. (a) What is its kinetic energy? (b) If the speed is dubled, what wuld be its kinetic energy? Slutin : (a) - (b)- K 1 1 = mν1 = (0.3)(15) = J K 1 1 = mν = (0.3)(30) = J Example : A mechanic pushes ( (ميكانيكي يدفع 500 kg car, mving it n frictinless flr frm rest with cnstant acceleratin. He des 5000 J f wrk in the prcess. During this time, the car mves 5 m. (a) What hrizntal frce did he exert n the car?, (b) what is the final speed f the car?
28 Slutin : W = Fd. cs θ = F. dcs = F. 5 F = 00 N 00 N = 500. a a = 0.04 m / s F = ma v = v + ax = 0 + (0.04)(5) = m / s Example : A frce F = (6i j) N acts n a particle that mves a displacement d = (3 i + j) m. Find (a) the wrk dne by the frce n the particle and (b) the angle between F and d. الشغل المبذول بالجاذبيةالا رضية Wrk dne by Gravitinal Frce The wrk dne by gravity can be given by W = F. d csφ The value f F is the gravitatinal frce, F g = mg. In case f rising an bject, F g is directed ppsite the displacement d, φ = 180 W = F. d cs φ = mg. d cs180 W = mg. d (1) When the bject is falling back dwn, the angle between F g and d is zer, φ = 0 W = F. d cs φ = mg. d cs 0 W = + mg. d () sign indicates that the gravitatinal frce transfer amunt f mg. d frm the kinetic energy f the bject during rising, while the + sign tells us that gravity add amunt f mg. d t kinetic energy f the bject.
29 If an external frce F is used t lift an bject, the applied frce tends t transfer energy t the bject while the gravitatinal frce F g tends t transfer energy frm it. The change in the kinetic energy f the bject K = Kf Ki = Wa W g 1-The Spring Frce الشغل المبذول بالزمبرك Wrk dne by Spring A blck n a hrizntal, frictinless surface is cnnected t a spring. If the spring is either stretched r cmpressed a small distance frm its unstretched (equilibrium) cnfiguratin, it exerts n the blck a frce f magnitude F x = kx Hke s law where x is the displacement f the blck frm its unstretched (x = 0) psitin and k is a psitive cnstant called " cnstant f the spring". The negative sign means that the frce exerted by the spring is always directed ppsite the displacement. Measuring spring cnstant : The spring is hung vertically. An bject f mass m is attached t its end. Under the actin f lad mg, the spring stretches distance d. Frce spring = weight f lad k. x= mg mg k = x
30 The wrk dne by the spring is given by W 1 k x =, where x is the change in the spring (القدرة ( Pwer المعدل الزمني لعمل الشغل Pwer is the time rate f ding wrk P dw dt = ( Jule/s = watt) P F. dx.csφ dx = = F.cs φ. d t dt P = Fv..csφ Example: A 0.6 kg particle has a speed f m/s at pint A and kinetic energy f 7.5 J at pint B. What is (a) its kinetic energy at A? (b) its speed at B? (c) the ttal wrk dne n the particle as it mves frm A t B? Slutin : (a) (0.6)() 1. KA = mv = = J
31 (b) (0.6) KB = mvb = vb = 7.5 = 0.3 v B (c) - W = W W = B A = = 6.3 J Prblem : 1- Express the unit f the frce cnstant f a spring in terms f the basic units meter, kilgram, and secnd. - when 4 kg mass is hung vertically n a certain light spring that beys Hke s law, the spring stretches.5 cm. If the 4 kg mass is remved, (a) hw far will the spring stretch if 1.5 kg mass is hung n it and (b) hw much wrk t stretch the same spring 4 cm frm its unstretched psitin? 3- A 15 kg blck is dragged ver a rugh, hrizntal surface by 70 N frce acting at 0 abve the hrizntal. The blck is displaced 5 m, and the cefficient f kinetic frictin is 0.3. Find the wrk dne by (a) the 70 N frce, (b) the nrmal frce, (c) the frce f gravity. (d) What is the energy lss due t frictin? (e) Find the ttal change in the blck s kinetic energy.
32 Ptential Energy As an bject falls tward the Earth, the gravitatinal frce des wrk n the bject and increases the bject s kinetic energy. The cnversin ( (التحول frm ptential energy t kinetic energy ccurs cntinuusly ver the entire fall. The prduct f the gravitatinal frce mg and height y f the bject gives the ptential energy (U g = mg.y). Fr either rise r fall, the change in the ptential energy U is equal t the negative f the wrk dne U = W U = mg.y mg.y 1 = U = mg (y y 1 ) In a blck-spring system, if we push the blck t left, the spring frce acts rightward and transfers kinetic energy f the blck t elastic ptential energy f the spring-blck system. The blck slws and stps and begins t mve rightward. The transfer f energy is reversed, frm elastic t kinetic energy. 1 U( x) = kx 1 1 U = kx f kxi = 1 U = k ( x f xi ) Cnservatin f Mechanical Energy The mechanical energy f a system is the sum f its ptential energy U and the kinetic energy K f the bject E mec = K + U Fr islated system which means that n external frce, frm utside the system, causes energy changes inside the system.
33 E = K + U = mec K = U K K = ( U U ) 1 1 K + U = K + U = cns tant 1 1 (قانون حفظ الطاقة الميكانيكية ( the This is called "Principle f Cnservatin f Mechanical Energy" 0 Example : A ball f mass m is drpped frm height h abve the grund. Determine the speed f the ball when it is at a height y abve the grund. Slutin : At the instant the ball is released, its kinetic energy is K i = 0 and the ptential energy f the system is U i = mgh. When the ball is at a distance y abve the grund, (kinetic energy + ptential energy) f = (initial kinetic + ptential energy) i Slutin : Ki + Ui = Kf + Uf mgh = mv f + mgy v = g ( h y) f Wrk Dne by an External Frce When yu lift a bk thrugh sme distance, the frce yu apply des wrk W n the bk, while the gravitatinal frce des wrk W g n the bk. The net wrk dne n the bk is related t the change in its kinetic energy as W + W g = K
34 Because the gravitatinal frce is cnservative, the wrk dne by the gravitatinal frce equals U. Substituting this int the abve equatin gives W = U + K The right side f the equatin represents the change in the mechanical energy f the bk Earth system. This result indicates that yur applied frce transfers energy t the system in the frm f kinetic energy f the bk and gravitatinal ptential energy f the bk Earth system. Example : A particle f mass m = 5 kg is released frm pint A and slides n frictinless track. Determine (a) the particle s speed at pints B and C and (b) the net wrk dne by the frce f gravity in mving the particle frm A t C. Slutin : Because the particle mves under cnservative frce KA + UA = KB + UB = KC + UC mgh = mv + mgh = mv + mgh gh v gh v gh A B B C C A = B + B = C + C (a) Substituting fr the height values at A, B, C by 5, 3., m, we can find ν A and ν B. (b) The net wrk dne by gravity = 0 Example : 3 kg mass starts frm rest and slides a distance d dwn a frictinless 30 incline. While sliding, it cmes int cntact with unstressed spring. The mass slides an additinal 0. m befre rest by
35 cmpressin f the spring (k = 400 N/m). Find the initial separatin d between the mass and the spring. Slutin : The frce acting n the blck F = mg sinθ = 5 x 9.8 x sin 30 = 4.5 N F = mg sinθ = m a a = g sinθ = 9.8 x sin 30 = 4.9 m/s = + v v ax v = 0 + (4.9)( d) = 9.8d 1 1 mv = k x 1 1 5(9.8 d ) = (400)(0.) 16 d = m 49
36 Rtatin Cnsider a flat and rigid bject f arbitrary shape rtating abut a fixed axis thrugh (ο). A particle at P is at a fixed distance r frm the rigin and rtates abut it in a circle f radius r. In plar crdinates, the pint P is given by (r, θ), where r is the distance frm the rigin t P and θ is measured cunterclckwise frm the psitive x axis. During rtatin, θ changes with time and r remains cnstant. (In cartesian crdinates, bth x and y vary in time). As the particle mves alng the circle frm the psitive x axis (θ = 0) t P, it mves thrugh an arc f length S, which is related t the angular psitin θ thrugh the relatinship θ = s r Because θ is the rati f an arc length and radius f the circle, it is a pure number. Cmmnly, we give θ an artificial unit radian (rad), where ne radian is the angle subtended by an arc length equal t the radius f the arc.
37 π θ( rad ) = θ (deg) 180 Fr example : 60 = π rad, 45 = π rad 3 4 In analgy t linear speed, the instantaneus angular speed ω is defined as d θ ω = dt Angular speed has units f radians per secnd (rad/s), r rather secnd s -1, because radians are nt dimensinal. The angular acceleratin is defined as dω d dθ α = = ( ) = d t dt dt d θ α = dt Angular psitin (θ), angular speed (ω), and angular acceleratin (α) are analgus t linear psitin (x), linear speed (v), and linear acceleratin (a). The variables θ, ω, and α differ dimensinally frm the variables x, v, and a nly by a factr having the unit f length. x = θ. r v = ω. r a = α. r Rtatin with Cnstant Angular Acceleratin As we cnsider equatins.1.5 are the basic equatins fr cnstant linear acceleratin, equatins are the basic equatins fr cnstant angular acceleratin. T slve any prblem, we chse an
38 equatin fr which the nly unknwn variable will be the variable requested in the prblem. υ= υ + at 1 x x = υt + at υ (.1) (.) υ = + ax ( x ) (.3) 1 x x = ( υ + υ ) t (.4) 1 x x = υt at (.5) ω = ω + αt 1 θ θ = ωt + αt (10.1) (10.) ω = ω + αθ ( θ) (10.3) 1 ( θ θ) = ( ω + ω) t (10.4) 1 θ θ = ωt αt (10.5) Example : A wheel rtates with a cnstant angular acceleratin f 3.5 rad/s. If the angular speed f the wheel is rad/s at t i = 0, (a) thrugh what angle des the wheel rtate in s? (b) What is the angular speed at t = s? Slutin : 1 θf θi = ωt + αt = 1 = ( rad / s )( s ) + (3.5 rad / s )( s ) = = 11 rad = 630 ω = ω + αt ω = ( rad / s ) + (3.5 rad / s )( s ) ω = 9 rad / s
39 Rtatinal Kinetic Energy Let us nw lk at the kinetic energy f a rtating rigid bject, cnsidering the bject as a cllectin f particles and assuming it rtates abut a fixed z axis with an angular speed ω. Each particle has kinetic energy determined by its mass and linear speed. If the mass f the i th particle is m i and its linear speed is v i, its kinetic energy is K 1 i = mv i i T prceed further, we must recall that althugh every particle in the rigid bject has the same angular speed ω, the individual linear speeds depend n the distance r i frm the axis f rtatin accrding t the expressin v i = r i ω. The ttal kinetic energy f the rtating rigid bject is the sum f the kinetic energies f the individual particles: 1 KR = Ki = mv i i = 1 KR = ( mi ri ω ) = 1 1 KR = ω mi ri = I ω where we have factred _ frm the sum because it is cmmn t every particle.
40
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