Chapter 6. Work and Energy

Transcription

1 Chapter 6 Wrk and Energy

2 6. Wrk Dne by a Cnstant Frce Wrk invlves frce and displacement. W = Fs N! m = jule ( J)

3 6. Wrk Dne by a Cnstant Frce

4 6. Wrk Dne by a Cnstant Frce Mre general definitin f the wrk dne n an bject, W, by a cnstant frce, F, thrugh a displacement, s, with an angle, θ, between F and s: W = ( F cs! )s cs0 cs90 cs80 = = 0 =!

5 6. Wrk Dne by a Cnstant Frce Example Pulling a Suitcase-n-Wheels Find the wrk dne if the frce is 45.0 N, the angle is 50.0 degrees, and the displacement is 75.0 m. ( ) [ F cs s = ( 45.0 N) cs 50.0 ]( 75.0 m) W =! = 70 J

6 6. Wrk Dne by a Cnstant Frce A weightlifter ding psitive and negative wrk n the barbell: Lifting phase --> psitive wrk. ( F cs ) s Fs W = 0 = Lwering phase --> negative wrk. ( F cs ) s = Fs W = 80!

7 Example. Wrk dne skatebarding. A skatebarder is casting dwn a ramp and there are three frces acting n her: her weight W (675 N magnitude), a frictinal frce f (5 N magnitude) that ppses her mtin, and a nrmal frce F N (6 N magnitude). Determine the net wrk dne by the three frces when she casts fr a distance f 9. m. W = (F cs θ)s, wrk dne by each frce W = (675 cs 65 )(9.) = 60 J W = (5 cs 80 )(9.) = - 50 J W = (6 cs 90 )(9.) = 0 J The net wrk dne by the three frces is: 60 J - 50 J +0 J = 470 J

8 6. Wrk Dne by a Cnstant Frce Example 3 Accelerating a Crate The truck is accelerating at a rate f +.50 m/s. The mass f the crate is 0 kg and it des nt slip. The magnitude f the displacement is 65 m. What is the ttal wrk dne n the crate by all f the frces acting n it?

9 6. Wrk Dne by a Cnstant Frce The angle between the displacement and the nrmal frce is 90 degrees. The angle between the displacement and the weight is als 90 degrees. Thus, fr F N and W, W ( F cs 90) = 0 = s

10 6. Wrk Dne by a Cnstant Frce The angle between the displacement and the frictin frce is 0 degrees. f s = ma = ( 0 kg)(.5m s ) = 80N W = [( ) ]( ) 4 80N cs0 65 m =.! 0 J

11 6. The Wrk-Energy Therem and Kinetic Energy Cnsider a cnstant net external frce acting n an bject. The bject is displaced a distance s, in the same directin as the net frce.! F s The wrk is simply W = (Σ F)s = (ma)s

12 6. The Wrk-Energy Therem and Kinetic Energy W = ( ) ( ) = m v! v = mv! mv m as f f v f = v + cnstant acceleratin kinematics frmula ( ax) ( ) ( ) ax =! v f v DEFINITION OF KINETIC ENERGY The kinetic energy KE f and bject with mass m and speed v is given by KE = mv

13 6. The Wrk-Energy Therem and Kinetic Energy THE WORK-ENERGY THEOREM When a net external frce des wrk n an bject, the kinetic energy f the bject changes accrding t W = KE! = mv! f KE f mv (als valid fr curved paths and nn-cnstant acceleratins)

14 6. The Wrk-Energy Therem and Kinetic Energy Example 4 Deep Space The mass f the space prbe is 474 kg and its initial velcity is 75 m/s. If the 56.0 mn frce acts n the prbe thrugh a displacement f m, what is its final speed?

15 6. The Wrk-Energy Therem and Kinetic Energy W = f mv! mv W = [(Σ F)cs θ]s

16 6. The Wrk-Energy Therem and Kinetic Energy [( ) ] F cs s = mv! mv [(Σ" F)cs θ]s # ( - ) ( 9 ) 5.60" 0 N cs0.4" 0 m = ( 474 kg) v! ( 474 kg)( 75m s) f f v f = 805m s

17 6. The Wrk-Energy Therem and Kinetic Energy Example. A 58 kg skier is casting dwn a 5 slpe as shwn. Near the tp f the slpe, her speed is 3.6 m/s. She accelerates dwn the slpe because f the gravitatinal frce, even thugh a kinetic frictinal frce f magnitude 7 N ppses her mtin. Ignring air resistance, determine the speed at a pint that is displaced 57 m dwnhill.

18 6. The Wrk-Energy Therem and Kinetic Energy v = 3.6 m/s s = 57 m 5 slpe m = 58 kg f k = 7 N Find v f

19 6. The Wrk-Energy Therem and Kinetic Energy W = mv! f mv Wrk-Energy Therem W = [(Σ F)cs θ]s " F = mg sin 5! f k = (58)(9.8)(0.4) - 7 = 70 N W = [(70)cs 0](57) = 9700 J Slving the Wrk-Energy Therem fr v f : v f = [(W + mv )/m] / = [((9700) + (58)(3.6) )/(58)] / = 9 m/s

20 6. The Wrk-Energy Therem and Kinetic Energy Cnceptual Example 6 Wrk and Kinetic Energy A satellite is mving abut the earth in a circular rbit and an elliptical rbit. Fr these tw rbits, determine whether the kinetic energy f the satellite changes during the mtin.

21 6.3 Gravitatinal Ptential Energy Find the wrk dne by gravity in accelerating the basketball dwnward frm h t h f. W = ( F cs! )s ( ) W gravity = mg h! h f

22 6.3 Gravitatinal Ptential Energy ( ) W gravity = mg h! h f We get the same result fr W gravity fr any path taken between h and h f.

23 6.3 Gravitatinal Ptential Energy Example 7 A Gymnast n a Trampline The gymnast leaves the trampline at an initial height f.0 m and reaches a maximum height f 4.80 m befre falling back dwn. What was the initial speed f the gymnast?

24 6.3 Gravitatinal Ptential Energy W = mv! f mv ( ) h! h = mv mg! f ( ) W gravity = mg h! h f ( h h ) v =! g! f v =! ( s )(.0 m! 4.80 m) = 8.40m s 9.80 m

25 6.3 Gravitatinal Ptential Energy W gravity = mgh! mgh f DEFINITION OF GRAVITATIONAL POTENTIAL ENERGY The gravitatinal ptential energy PE is the energy that an bject f mass m has by virtue f its psitin relative t the surface f the earth. That psitin is measured by the height h f the bject relative t an arbitrary zer level: PE = mgh N! m = jule ( J)