Knetc Energy (energy of moton) E or KE K = m v = m(v + v y + v z ) eample baseball m=0.5 kg ptche at v = 69 mph = 36.5 m/s K = mv = (0.5)(36.5) [kg (m/s) ] Unts m [kg ] J s (Joule) v = 69 mph K = 00 J v = 00 mph K = 0 J!!! Lethal energes 5/6-
Work one by force F actng over splacement F F OR F F W= F W= F component (projecton of) component (projecton of) F along along F W= F cos() Unts m m (N) m (Kg ) m Kg s s J!! 5/6-
F W= -F eample N mg Collnear F an F W= +F bo on frctonless plane T N mg T T =T cos() W tot =T =T cos() Note knetc frcton force always opposes splacement always oes - work Constant force not // note 5/6-3
E: object lowere on a strng T F = T - mg h m g = h Work one by gravty W g = mgh (postve work) Total Work Done on Object Work one by tenson W T = -Th (negatve work) W = (T - mg) h W = W g - W T Net Work If object lowere wth constant velocty, then mg - T = 0 an W =0 If object accelerates own, then 0 < W < mg 5/6-4
Work-Energy Theorem Object: m; constant force F; constant accelertaton a=f/m v - v a= 0 v + v v = v 0 + at 0 = [ ] t t W = F = ma v - v v + v 0 0 W = m ( ){[ ] t} t m W = (v - v 0 ) mv = K = Knetc Energy 0 true for non-const. a also!!! W = mv - mv0 W = K - K W = K Total work one on object= change n knetc energy (work one by total force) m [K] = kg ( s ) Joule = Nm or 5/6-5 ft [K] = slug ( s ) ft lb
Power A measure of the rate at whch work s one. Average Power P = W t Instantaneous Power W P = t Power= work per unt tme [ J/s= Watt] SI unt: J/s = watt, W horsepower = hp = 746 W 5/6-6
0 N F 5 kg m 3m Eample W = F = 0 (3) Nm = 30 J Assume: no other forces (no frcton) & starts from rest (v = 0) W= K What s v f? W= F = ½ mv f - ½ mv 30 J = ½ (5 kg) v f v f = 3.5 m/s If ths took 3 sec what was the average power nput by F? P = W/ t = 30 J/3s = 0 watts 5/6-7
Object projecte up nclne plane wth spee v 0 How far up? Work Energy Theorem v 0 N Normal force oes no work N // = 0 : N to W N = N = 0 - mv 0 = -mg sn() = V 0 gsn W g = F g W g = (-mgsn) W W W mv mv N g f 0 5/6-8 F g mg - F g sn -mgsn Note: h=heght up plane h= sn() h - mv 0 = -mgh
Conservatve Force n general 3 ways to efne a conservatve force F = conservatve force f:. the work one by F n any roun trp moton of an object s zero W F = 0 Path. the change n knetc energy, K, cause by F n any roun trp moton of an object s zero. K F = 0 Path 3. the work one by F when an object moves from an ntal pont to a fnal pont epens only on these two ponts an not on the path taken between them. Path conservatve force eamples: - Gravty - Electrc fels - The force from a sprng
Work Energy Theorem K W f K - Wf Potental Energy & Conservaton of Energy 0 K knetc energy change Now efne change n Potental Energy U U - potental energy change 0 * W f * or knetc energy for conservatve forces K Energy conservaton E E f U E potental energy ntal & fnal postons/tmes 5/6-7 * E conserve total mechancal energy
E K U E E f Energy conservaton In the case of gravtatonal force/potental (near the earth's surface) E knetc energy change U = mg (y-y o ) mv potental energy change mg (y - y o ) y o s some arbtrary poston where U=0 Can choose zero of potental anywhere one wants - once choce mae keep same 5/6-8 energy conserv60-0-.ppt
Eample y= h Throw an object up how hgh oes t go? E E f mv v top = 0 : know E f 0 mgh mgh h h v g h 00 (9.8) 50 m m s m ( ) s U = 0 y = 0 v = 00 m/s E mv 0 5/6-9
5/6-0
Eample roller coaster starts from rest at top of st hll h o h E = mv o + mg h o E = mv + mg h E 3 = mv 3 + mg h3 0 but h 3 = h0 E = mg h o E 3 = mv 3 + mg h0 E = E E = E 3 mg h o = mv + mg h mg h 0 = mv 3 + mg h0 mv = mg (h -h o ) 0= mv3 0= v 3 v = g (h-h o) Object comes to rest before top of 3 r hll (when h=h 0 ) 5/6- h 3
Energy conservaton fast way to solve for h vs v 5/6-0a
Ball thrown upwar near surface of the earth y= h y = 0 W = F mgh mg y= h y = 0 mg W = F mgh mg y= h y = 0 mg W = F mgh Inee, n every case you nvent the work one by gravty wll epen only on the vertcal change n heght (h) an K=mgh!!!! 5/6- All these cases have the same the =h an the work one by gravty s the same = mgh!!!! In all these cases (by the W-E theorem) the change n knetc energy K=mgh!!!! Gravty s a very epenable conservatve force
Eample: loop the loop conservaton of energy s great but on t forget Newton s Law s!!! Q. How hgh oes h have to be to have v go to 0 At top of loop? (wrong queston f you want to survve!!) A. h=r But you fall off before you get there!! 5/6-
loop the loop 5/6-a v s mn at top of loop at top of loop v N mg v (ncreasng) v (ecreasng) 5/6-a v (ma)
k k - k - - k F W U=0 at =0,.e. = 0 k k - W k U k mv U K E k k W - U F -k Sprng: Energy Conseratons the larger the splacement the larger the restorng force Hooks Law Force restores to equlbrum (=0) 5/6-3
Energy cyclc transfer :: knetc energy potental energy = 0 = A = -A E E E mv ma ka ka E ka mv v ma A mv k m ma k 5/6-4
Now nclue non-conservatve forces n Work Energy Theorem Work Energy Theorem W + W + + W n conservatve forces force - U - U - - U n + W nc = K changes n potental energes assocate wth conservatve forces Work one by W nc = E non-conservatve forces + W nc = K W nc = K + U + U + + U n 5/6-5 Work one by non-conservatve change n total mechancal energy change n KE non-conservatve forces knetc frcton (-work) rocket, eploson, kck (+work)
m =.5 kg v k = 30 N/m non-conservatve force eample = 0 k =0.5 f f = 0.075 m v=? NC frcton, - work (splacement) [force] W f =E f -E [-μ mg]( )= k - mv k f f mv kf μkmg k v μ g m f k f 30 m (0.075) (.5)9.8(0.075).5 s m v=.04 s W = -μ mg f k 5/6-6
5/6-7 non-conservatve force eample
non-conservatve force eample Conser an elevator that falls from a heght h, but whch s brought to rest by the brakes just as t reaches the groun free fall brakng h v F g = mg W f = E = mgh f F f only acts over then F f = frctonal brakng force Velocty before brake ½ mv = mg(h-) F g = mg W f = E = mgh = F f F f = mgh/ 5/6-7a
y= h y = 0 v up v f Ball thrown upwar wth v 0 mg W up =-mg h= - mv own W own =+mg h= Also follows from energy conservaton v 0 mv f W roun trp = W up + W own = 0 0 Recall W = F = mv - mv v 0 K roun trp = 0 v f = f W =K mg W = F 5/6-8
A note on a why a weght hangng vertcally from a sprng can be treate lke one on a frctonless surface. Appen 5/6-App9
+ T system Inclne Plane + a Pulley N T What s a of objects? + m gsn T s an nternal force m m g m gcos to plane N - m gcos = 0 system F = ma (along the recton of potental moton) - m g + m gsn = (m + m )a Tot eternal force on system Revew peoblem Tot mass of system 5/6-0 a = g m sn - m m + m