Physics 111. Lecture 18 (Walker: 8.3-4) Energy Conservation I March 11, Conservation of Mechanical Energy

Transcription

1 Physics 111 Lecture 18 (Walker: 8.3-4) Energy Conservation I March 11, 2009 Lecture 18 1/24 Conservation o Mechanical Energy Deinition o mechanical energy: (8-6) I the only work done in going rom the initial to the inal position is done by conservative orces: Or equivalently: Lecture 18 2/24

2 Conservation o Mechanical Energy Mechanical energy conservation can make kinematics problems much easier to solve: U i =mgh K i = 0 U i =0 K i = 0 U =0 K =½mv 2 U =-mgh K =½mv 2 Lecture 18 3/24 Conservation o Mechanical Energy Emech = K + U Emech = K + U = 0 (Conservation o Mechanical Energy) Note that K U; the correct statement is K= U Lecture 18 4/24

3 Example Problem Tarzan swings on a 30.0-m-long vine initially inclined at an angle o 35.0 with the vertical. What is his speed at the bottom o the swing (a) i he starts rom rest? (b) i he pushes o with a speed o 4.00 m/s? 30 m Lecture 18 5/24 Choose the zero o potential energy at bottom o arc. Tarzan s initial height above this level is: y i = (30m)[1-cos(35 )] = 5.4 m Two orces act during the swing, gravity (conservative) and vine tension (not conservative, but does no work). Thus E = E i. y i y a) ½mv = ½mv i2 + mgy i = 0 + mgy i v 2 = 2gy i = 2(9.8m/s 2 )(5.4m) and thus v = 10.3 m/s b) ½mv = ½mv i2 + mgy i v 2 =v i2 +2gy i =(4.0m/s) 2 +2(9.8m/s 2 )(5.4m); v =11.0 m/s Lecture 18 6/24

4 Question When a ball o mass m is dropped rom height h, its kinetic energy just beore landing is K. I a 2 nd ball o mass 2m is dropped rom height h/2, what is its kinetic energy just beore landing? m h 2m (a) K/4 (b) K/2 (c) K (d) 2K (e) 4K Lecture 18 7/24 Strategy: Conservation o Mechanical Energy MODEL: Choose a system without riction losses or other losses o mechanical energy. VISUALIZE: Draw a beore-and-ater pictorial representation. Deine symbols that will be used in the problem, list known values, and identiy what you re trying to ind. SOLVE: The mathematical representation is based on the law o conservation o mechanical energy. ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Lecture 18 8/24

5 What is the Final Speed? v=0 v=4 m/s v = 1 m/s, 2 m/s, or 3 m/s? v=5 m/s Lecture 18 9/24 Energy is a scalar. Speed o cap is v i at height y i and v at height y, independent o the path between the two heights. Angle at which cap is launched does not matter, as long as v i is large enough to carry cap to height y. Speed and Path Lecture 18 10/24

6 Example: Catching a Home Run Player hits 0.15 kg baseball over outield ence. Ball leaves bat with speed o 36.0 m/s and a an in bleachers catches it 7.2 m above the point where it was hit. Neglect air resistance. (a) What is the kinetic energy K o the ball when caught? (b) What is the speed v o the ball when caught. Lecture 18 11/24 U E = U + K = mgy + mv 1 2 i i i 2 i = mgy = = 2 (0.15 kg)(9.81 m/s )(7.2 m) 11 J K = E U = (97 J) (11 J) = 86 J v = 0 + ½(0.15 kg)(36 m/s) 2 = 97J 2K 2(86 J) = = = 34 m/s m (0.15 kg) Lecture 18 12/24

7 How large does h need to be so block doesn t all o track at top o circle? (Frictionless track.) At top o circle, want mg = mv t2 /r (Why?) Thus need v t2 = rg Lecture 18 13/24 E t = E h ½mv t2 + mgy t = ½mv h2 + mgy h ½m(rg) + mg(2r) = 0 + mgh h = 2r+½r = 2.5 r Lecture 18 14/24

8 Example: Spring Time Block o mass m slides on horizontal, rictionless surace until it hits a spring with orce constant k. Block comes to rest ater compressing spring a distance d. Find initial speed o block. (Ignore air resistance and energy lost when block initially hits spring.) Lecture 18 15/24 Only the spring orce does work, and it is conservative. Thus E = E i ½kx 2 + ½mv 2 = ½kx i2 + ½mv i 2 ½kd = 0 + ½mv i 2 v i2 = v i = d kd 2 m k m Lecture 18 16/24

9 Basic Energy Model 1. Kinetic energy K is associated with motion o a particle. Potential energy U is associated with position o particle when conservative orce acts. 2. Kinetic energy can be transormed into potential energy, and potential energy into kinetic energy. 3. Under some circumstances, mechanical energy E= K + U is conserved -- its value at end o a process is same as at beginning. Q1: Under what circumstances is E conserved? Q2: What happens to the energy when E is not conserved? Q3: How do you calculate U or orces other than gravity? Lecture 18 18/24

10 Nonconservative Forces In the presence o nonconservative orces, the total mechanical energy is not conserved: Solving, (8-9) Lecture 18 19/24 Example: Judging a Putt Goler badly misjudges a putt, giving the ball an initial speed v 1 which sends the ball a distance d that is only one quarter o the distance to hole. I nonconservative orce F due to resistance o grass is constant, what initial speed v 2 is needed to putt ball rom initial position to hole? E = K K = 0 mv = Fd i E = mv = F(4 d) 2 v = 2v Lecture 18 20/24

11 Example: Landing with a Thud Block o mass m 1 = 2.40 kg is on horizontal table with a coeicient o riction µ k = and is connected to hanging block o mass m 2 = 1.80 kg. When blocks are released, they move a distance d = 0.50 m, and then m 2 hits the loor. Find speed o the blocks just beore m 2 hits. Lecture 18 21/24 U = mgh+ mgd; K = 0; E = mgh+ mgd i 1 2 i i = = U mgh mg(0); K mv mv ; E E = m 1 gh + ½m 1 v 2 + ½m 2 v 2 E = E E = m + m v m gd nc W nc = - k d= -µ k m 1 gd i 1 2 ( 1 2) 2 2 E = W m + m v m gd = µ m gd v 2gd m 1 2 ( 1 2) 2 2 k 1 ( µ m ) 2(9.81 m/s 2 )(0.50 m) [(1.80 kg) (0.45)(2.40 kg) ] 2 k 1 = = = m + m (2.40 kg) + (1.80 kg) 1 2 = 1.30 m/s Lecture 18 22/24

12 Lecture 18 23/24 End o Lecture 18 Beore Friday, read Walker Homework Assignments #8b should be submitted using WebAssign by 11:00 PM on Friday, Mar. 13. Lecture 18 24/24