Problem 07-50 A 0.25 kg block s dropped on a relaed sprng that has a sprng constant o k 250.0 N/m (2.5 N/cm). The block becomes attached to the sprng and compresses t 0.12 m beore momentarl stoppng. Whle beng compressed, A) What s the work done on t b grat? B) What s the work done on t b the sprng orce? C) What s the speed o the block just pror to httng the sprng?
Problem 07-50 A 0.25 kg block s dropped on a relaed sprng that has a sprng constant o k 250.0 N/m (2.5 N/cm). The block becomes attached to the sprng and compresses t 0.12 m beore momentarl stoppng. Whle beng compressed, A) What s the work done on t b grat? B) What s the work done on t b the sprng orce?
Problem 07-50 A 0.25 kg block s dropped on a relaed sprng that has a sprng constant o k 250.0 N/m (2.5 N/cm). The block becomes attached to the sprng and compresses t 0.12 m beore momentarl stoppng. C) What s the speed o the block just pror to httng the sprng?
Force F (3.0î 7.0ĵ 7.0kˆ) Problem 07-60 n t 4.0s Intal Dsplacement d (3.0î 2.0ĵ 5.0kˆ) Fnal Dsplacement d ( 5.0î 4.0ĵ 7.0kˆ) A) Work done on the partcle B) Aerage power C) Angle between d and d
Force F (3.0î 7.0ĵ 7.0kˆ) Problem 07-60 n t 4.0s Intal Dsplacement d (3.0î 2.0ĵ 5.0kˆ) Fnal Dsplacement d ( 5.0î 4.0ĵ 7.0kˆ) A) Work done on the partcle W F s constant n ths problem, so F ( d d ) (3.0î 7.0ĵ 7.0kˆ) (( 5.0 3.0)î ((4.0 (3.0î 7.0ĵ 7.0kˆ) ( 8.0î 6.0ĵ 2.0kˆ) 24.0 42.0 14.0 32.0J 2.0)ĵ (7.0 5.0)kˆ)
Force F (3.0î 7.0ĵ 7.0kˆ) Problem 07-60 n t 4.0s Intal Dsplacement d (3.0î 2.0ĵ 5.0kˆ) Fnal Dsplacement d ( 5.0î 4.0ĵ 7.0kˆ) B) Aerage power P ae W/ t 32.0J/4s 8.0W
Force d F (3.0î 7.0ĵ 7.0kˆ) Intal Dsplacement (3.0î 2.0ĵ 5.0kˆ) d 38 Problem 07-60 n t 4.0s Fnal Dsplacement d d ( 5.0î 90 4.0ĵ 7.0kˆ) C) Angle between d and d d and d are two ectors, so take the dot product d d cosφ d d ( 15.0 8.0 35.0kˆ) / φ cos ( 5.0î cosφ 1 4.0ĵ 7.0kˆ) (3.0î 2.0ĵ 5.0kˆ) / 0.205 ((9.49)(6.16)) 78.2 o 12.0 / 58.5 ( d d ) 0.205
Problem 07-70 m 230kg, L 12.0m arable horzontal orce F horzontal dstance d 4.0m θ A) magntude o F n nal poston B) Total Work done on crate C) Gratatonal Work D) Work done b rope E) Work done b orce F
Problem 07-70 m 230kg, L 12.0m arable horzontal orce F horzontal dstance d 4.0m θ
Problem 07-70 m 230kg, L 12.0m arable horzontal orce F horzontal dstance d 4.0m θ F T sn θ mg tan θ
Dal Quz, September 21, 2004 Whch content do ou preer or the Tuesda lectures wthout the scheduled eams? 1) Lecture oer materal as we e done so ar. 2) Concentrate on workng problems and eamples. 3) No Lecture, just hold normal oce hours.
Drectons are Important n Space! Vector (drecton mportant) Quanttes: dsplacement eloct acceleraton Scalar (drecton not mportant) Quanttes: dstance speed acceleraton (same word, but t reall s a ector)
a a Components o Vectors a cosθ a sn θ Component notaton s magntude-angle notaton a a 2 a 2 tan θ a a
Add ectors b components )kˆ b (a )ĵ b (a )î b (a b a r kˆ b ĵ b î b b kˆ a ĵ a î a a z z z z r r r r r
Scalar product o two ectors a and b a.b a b cos φ a, b: magntude o a, b φ: angle between the drectons o a and b Scalar product r a a î r b bî r r a b (a a b b ) (a ĵ ĵ b ) a b z z kˆ kˆ (a z b z )
Dsplacement s the change n poston (or locaton) 2-1 Dsplacement Dsplacement s a ector wth both magntude and drecton
Instantaneous eloct Veloct at a gen nstant V lm t->0 ( / t) d /dt the slope o (t) cure at tme t the derate o (t) wth respect to t VECTOR! -- drecton and magntude (Instantaneous) speed s the magntude o the (nstantaneous) eloct
Acceleraton Acceleraton s the change n eloct The aerage acceleraton 2 1 a ag t t t 2 (Instantaneous) acceleraton d d d d a a ( ) dt dt dt dt Unt: m/s 2 It s correctl a ector! 1 2 d 2 dt
Projectle Moton Acceleraton s constant and onl acts n the drecton The horzontal moton and the ertcal moton are ndependent o each other Horzontal moton: (Moton wth constant eloct) 0 0 cosθ 0 0 ( 0 cosθ 0 ) t Vertcal moton: (Moton o ree-allng object) 0 a t ( 0 snθ 0 ) g t 0 0 t ½ a t 2 ( 0 snθ 0 ) t ½ g t 2 assume the upward drecton s poste
Force F Chapter 5 Force and Moton s the nteracton between objects s a ector causes acceleraton Net orce: ector sum o all the orces on an object. N Ftotal Fnet F F1 F2 F3 F4 1... ma
Newton s Laws o Moton Newton s rst law: I no orce acts on a bod, then the bod s eloct cannot change, that s, the bod can not accelerate rest, stll rest mong, contnue mong wth same eloct Newton s second law: The net orce on a bod s equal to the product o the bod s mass and the acceleraton o the bod: Σ F m a Newton s thrd law: When two bodes nteract, the orces on the bodes rom each other are alwas equal n magntude and opposte n drecton: F AB F BA
Newton s Second Law Newton s second law: The net orce on a bod s equal to the product o the bod s mass and the acceleraton o the bod Σ F m a Σ F: ector sum o all the orces that act on that bod Σ F m a Σ F m a Σ F z m a z Unt: 1 N (1kg).(1m/s 2 ) 1 kg.m/s 2
A block les on a horzontal loor. a) What s the magntude o the rcton orce () on t rom the loor? 0N, but note S,ma µ S F N b) I a horzontal orce o 5 N s now appled to t, but t does not moe, what s now? s 5N c) I s, ma 10 N, wll the block moe the horzontal appled orce s 8 N? d) How about 12 N? Frctonal Forces es, because F > s no, because F < s S,ma µ S F N F N mg F mg
Unorm crcular moton Perod o reoluton (perod) T 2πr/ Centrpetal acceleraton magntude: a 2 /r drecton: radall nward and a: constant magntude but ar contnuousl n drecton Centrpetal orce: F ma m 2 /r ( r)
Work-Knetc Energ Theorem The change n the knetc energ o a partcle s equal the net work done on the partcle K K K 1 2... or n other words, m 2 1 2 m 2 W net r F dr nal knetc energ ntal knetc energ net work 1 2 1 2 K m K Wnet m 2 2 r W net
Work done b a arable orce W j F j, ag W Σ W j ΣF j, ag lm 0 ΣF j, ag W F()d Three dmensonal analss W r r F d F dr F d z z F dz z
Power The tme rate at whch work s done b a orce Aerage power P ag W/ t (energ per tme) Instantaneous power P dw/dt (Fcosφ d)/dt F cosφ F. Unt: watt 1 watt 1 W 1 J / s 1 horsepower 1 hp 550 t lb/s 746 W klowatt-hour s a unt or energ or work: 1 kw h 3.6 M J