Physics 231 Lecture 9

Transcription

1 Physics 31 Lecture 9 Mi Main points o today s lecture: Potential energy: ΔPE = PE PE = mg ( y ) 0 y 0 Conservation o energy E = KE + PE = KE 0 + PE 0

2 Reading Quiz 3. I you raise an object to a greater height, you are increasing A. kinetic energy. B. heat. C. potential energy. D. chemical energy. E. thermal energy. Slide 10-10

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4 Checking Understanding Three balls are thrown o a cli with the same speed, but in dierent directions. Which ball has the greatest speed just beore it hits the ground? A. Ball A B. Ball B C. Ball C D. All balls have the same speed Slide 10-6

5 Potential energy For certain orces, the work done by the orce in going rom position (x 0,y 0 ) to position (x,y) depends only the displacement and not on the path taken. Such a orce is called conservative. gravity is such a orce. Consider the work on a mass m under a displacement Δs at an angle θ with respect to the vertical as shown below: Δs (x 0,y 0 ) θ (x,y) W gravity F g = mg downwards g ( ) Δs = mg( y y) = Fgravity cos θ 0 From work - energy theorem : mg ( y0 y ) = KE KE0 From this we can see that being higher initially means that you can have a higher inal kinetic energy. Thus, mg(y 0 -y) is the part o the stored potential energy, which was changed into kinetic energy as the object moves rom its initial to its inal position. The potential energy only depends on the dierence in height between the initial and inal positions, i.e. on the vertical component o the displacement.

6 Example What dierence between the PE o a 000 kg car raised 10 m in the air and that o the same car on the ground? How much work would it require to lit it to that height? Δ PE = mgh = (000kg)(9.8m / s )(10m) Δ PE = x10 J 5 W =Δ PE= 1.96x10 J The potential energy o the car is 1.95x10 5 J larger when the car is 10 m above the ground than it is on the ground. It would take 1.95x10 5 J o work to lit it the 10 m.

7 Gravitational potential energy The most important t point o a potential ti energy is that t is only a unction o the position and not o the path taken to get there. I we break the path o the pail in to vertical sections the potential energy change is just mgδy and horizontal sections where the potential energy remains constant, we can see that the potential energy depends on the total vertical displacement and is independent o the path over which it is achieved. PE-PE 0 =mg(y-y 0 ) (x,y) (x 0,y 0 ) Thus we can deine PE=0 or some y 0 and then PE=mg(y-y 0) thereater. PE is W grav. It is the work one would need to do to move the pail rom A to B.

8 Conceptual problem At the bowling alley, the ball-eeder mechanism must exert a orce to push the bowling balls up a 1.0-m long ramp.the ramp leads the balls to a chute 0.5 m above the base o the ramp. Approximately how much orce must be exerted on a 5.0-kg bowling ball? a) 00 N b) 50 N c) 5 N d) 5.0 N e) impossible to determine Hint: the orce is conservative. It doesn t matter how you get up there. work = PE = FΔs PE mgδy F = = Δs Δs ( ) ( ) ( ) J = 5N 1m

9 Conservation o energy Up to an additive constant, we can deine PE=mgy. It is equal in magnitude but opposition in sign to the work being done by gravity.. Then KE KE0 = PE0 PE We can deine the total energy E as the sum o kinetic and potential energy. Then we have E = KE + PE = KE 0 + PE 0 Thus, when all work being done by orces in a problem can be expressed in terms o a potential energy, the total energy is conserved (i.e. remains constant). This is true or gravity and many other orces, but not or riction, or example.

10 Checking Understanding Three balls are thrown o a cli with the same speed, but in dierent directions. Which ball has the greatest speed just beore it hits the ground? A. Ball A B. Ball B C. Ball C D. All balls have the same speed E= KE + PE = KE + PE E= KE = mv = KE + PE PE 0 0 Slide 10-6

11 Example A water slide is constructed t so that t swimmers, starting ti rom rest at the top o the slide, leave the end o the slide traveling horizontally. As the drawing shows, one person is observed to hit the water 5.00 m rom the end o the slide in a time o s ater leaving the slid. Ignoring riction and air resistance, ind the height H in the drawing. Δx 5 m Δt 0.5 s h? v 0 0 v? H? Δx x 5m Δ x = vt; v = = = 10m/s t 0.5s h = height at bottom o slide: 1 -h = gt h = (4.9m 9m/s) 0.5s = 1.3m ( ) Conservation o energy (deine PE=0 at h): 1 mv + 0 = 0 + mg H h v H= h+ = 6.33m g ( )