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1 Physics 101 Sectio 3 Mar. 1 st : Ch. 7-9 review Ch. 10 Aoucemets: Test# (Ch. 7-9) will be at 6 PM, March 3 (6) Lockett) Study sessio Moday eveig at 6:00PM at Nicholso 130 Class Website: Or go directly to my website

ΔK Wet = F et ds 1 K = mv Mechaical eergy

2 Work ad Eergy System Work doe by o-coservative forces Work doe by coservative forces Wcoser. Potetial eergy Work-eergy theorem W et = ΔK Wet = F et ds 1 K = mv Mechaical eergy Work doe E K + U = ΔU Δ E =Δ K+ΔU 1 Ugrav. ( y) = mgy; Uelas. ( x) = kx

3 Work ad Eergy Coservatio of mechaical theorem System E = K + U = mec Δ E = 0 mec Cost. More tha mechaical eergy ΔE Δ E +Δ E +ΔE mec themal iteral System W =Δ E

4 Work ad Eergy W = Δ E = 0 Isolated system Coservatio of total eergy: The total eergy of a isolated system caot chage

5 Impulse ad Mometum J = Δ P= Pf Pi P mv ( P m v ) = = t f J = t i Fdt System Ay exteral impulse icreases the mometum of a system Nothig but Newto d -law F = ma = t f J Fdt P P i = = f t i dp dt

6 Impulse ad Mometum Impulse from exteral forces J ext. Impulse from iteral forces J it eral = 0 (collisio forces) F = ext. 0 Isolated System Coservatio of total Mometum: The total mometum of a isolated system caot chage Δ P= Δ m v = 0

7 Impulse ad Mometum F ext. = 0 Applicatio to collisio: Δ P= Δ m v = 0 Elastic collisio: Δ P= Δ m v = 0 1 Δ K = Δ m v = ( ) 0 The total mometum ad total kietic eergy caot chage

8 Frictio Force Problems E = K + U K + U = E f f f i i i

9 No-coservative Forces: Frictio Mechaical eergy: Coservative System Emech = K+ U Frictio takes Eergy out of the system treat it like thermal eergy ΔE mech 0 Lose Mechaical eergy I geeral lthe work kdoe by a exte ral lforce is W= ΔE= Δ E +ΔE mech Thermal If there is o work doe byaexteral force ΔE mech +ΔE Thermal = 0 E ( fial) ) = E (iitial) mech E ( fial) ) mech th + E th (iitial) K f +U f = K i +U i F f displacemet ΔE Thermal

10 A ew way to look at Frictio Questio #8: A block of mass m slides dow the iclied plae startig with zero velocity. Regio D has frictio ad it comes to rest after movig a distace D. Mechaical Eergy is ot Coserved. E mec ( fial) = E mec (iitial) ΔE th E mec ( fial) = E mec (iitial) F f x We ca draw this. E mec ( fial) = E mec (iitial) ΔE th E mec ( fial) = E mec (iitial) F f x K = mv

11 A ew way to look at Frictio Questio #9: Fid K ad E mec as the block moves alog. E mech -F f distace K = mv

12 Problem: No Frictio The pulley is massless ad the iclie is frictioless. If the blocks are released from rest with the coectig cord taut, block B accelerated dowward. What is their total kietic eergy whe block B has falle a distace L? From coservatio of Mechaical Eergy, the chage i total potetial eergy is equal AND opposite (i.e. egative) to the chage i the total kietic eergy. Iitially the blocks are statioary, so that KE i =0 We defie the potetial eergy to be zero iitially. This meas E mech = 0 Whe Block B has falle L m, Block A has rise [(L)siθ] Thus, the fial total kietic eergy is: E mec = K f +U f = K i +U i = ( K f +U f = m A + m B )v m B gl + m A Lgsiθ = 0 ( K f = m A + m B )v = m B gl m A Lgsiθ

Problem: Frictio The pulley is massless ad the iclie has a kietic coefficiet of frictio μ k. If the blocks are released from rest with the coectig cord taut, block B accelerated dowward.

Iitially the blocks are statioary, so that K i =0 We defie the potetial eergy to be zero iitially.

13 Problem: Frictio The pulley is massless ad the iclie has a kietic coefficiet of frictio μ k. If the blocks are released from rest with the coectig cord taut, block B accelerated dowward. What is their total kietic eergy whe block B has falle a distace L? Not a coservative force so Mechaical Eergy is ot coserved. Iitially the blocks are statioary, so that K i =0 We defie the potetial eergy to be zero iitially. This meas E mech = 0 Whe Block B has falle L m, Block A has rise [(L)siθ]: U f = -m B gl+m A glsiθ The Eergy removed (thermal) is the work doe by frictio = m A glcosθ K f +UU f = K i +UU i ΔE th = ΔE th ( K f +U f = m + m A B)v m B gl + m A Lgsiθ = m A glμ k cosθ ( K f = m + m A B)v ( ) ( ) ( ) v == Lg m m siθ + μ cosθ B A k ( m A + m B ) = m B gl m A Lg siθ + μ k cosθ

14 A 10 kg block is released from poit A i the figure. The track is frictioless except for the portio betwee poits B ad C, which has a legth of 6 m. The block travels dow the track, hits a sprig of force costat 50 N/m, ad compresses the sprig 0.3 m from its equilibrium positio before comig to rest mometarily. Determie the coefficiet of Kietic frictio betwee the block ad the rough surface betwee B ad C. Lets draw the picture agai! E mec (iitial) = mgh Eergy lost=f f d = mgμ k d After the block passes the frictio surface So whe it compresses the sprig v=0 kx E mech = K +U = mv = mgh mgμ kd = mgh mgμ d k

15 Problem: Frictio A block slides alog a track from oe level to a higher level, by movig through a itermediate valley. The track is frictioless util the block reaches the higher level. There is a frictioal force that stops the block i the distace d. The block s iitial speed is v 0 ; the height differece is h; the coefficiet of kietic frictio is μ k. What is d? What do we kow? Alog the part of the track which is frictioless, coservatio of Mechaical Eergy holds. However, at the top the frictio trasfers eergy out of the system (ΔE therm ). Sice the system is isolated the chage i the total eergy is zero. [ ] 0 = Wext, et = Δ K +Δ Uet +Δ Etherm 0 = 0 1 mv [( 0)+ ( mgh 0) ]+ ( μ k ( mg) )d d = 1 1 μ k mg mv 0 mgh Solvig for d = v 0 μ k g h μ k ( )

16 Problem 8-63 A particle ca slide alog a track as show. The curved portios of the track are frictioless, but the flat part has a coefficiet of kietic frictio of μ k = 0.0. The particle is released from rest at poit A, which is a height h=l/ above the flat part of the track. Where does the particle fially stop? Iitial mechaical eergy 1 E mec = U grav = mgh = mgl Eergy lost to frictio every time goig through the flat part W = μ mg L fric k Assume the particle moves trips (ot ru-trip) Δ E = W = W fric 1 mgl = μkmg L 1 = = 5.5 μ k 0.5mgL 0 Stop at the middle of the flat part!

PROBLEM Copyright McGraw-Hill Education. Permission required for reproduction or display. SOLUTION. v 1 = 4 km/hr = 1.

PROBLEM Copyright McGraw-Hill Education. Permission required for reproduction or display. SOLUTION. v 1 = 4 km/hr = 1. PROLEM 13.119 35,000 Mg ocea lier has a iitial velocity of 4 km/h. Neglectig the frictioal resistace of the water, determie the time required to brig the lier to rest by usig a sigle tugboat which exerts