Conservation of Energy

1. Conservation of Energy If a body is acted upon by a force that does no total work during any round trip/cycle A, then the force is conservative. Otherwise, it is non-conservative B! A) gravity: DOWN: W = mgh UP: W = - mgh Wnet = 0 conservative!

2. A) spring: W = +.5kd2 Return trip is exact opposite: Wnet = 0 W = - .5kd2 conservative

3. Work done in a conservative system is path independent– like in the case of gravity!

4. B) Friction: The total work done by friction during a complete cycle is not 0! Friction is a non-conservative force!

5. Potential Energy (U) Potential Energy is the energy stored within a system-- also called the energy of configuration. • it can only be defined in terms of conservative forces-- not non-conservative forces! If we are dealing with a conservative system, then the change in Potential Energy will be equal to the work done by the conservative force: ∆U = -W

6. If the system is conservative, then the total amount of energy will remain a constant value: U + K = E This is the mathematical representation of the law of conservation of mechanical energy. Therefore: ∆U + ∆K = 0 This will hold true for more than one conservative force (more than one force doing the work) as well!

7. In the Conservative system: ∆U + ∆K = 0 In the non-conservative (and conservative) part of the system: W = ∆K Assume a conservative system unless otherwise specified!

8. A spring gun used to horizontally launch a 12 g ball is compressed 3.2 cm from its relaxed state and loaded. Use the law of conservation of energy to find the speed of the ball as it leaves the gun if k for the spring is 7.5 N/cm. ∆U + ∆K = 0 v = d k m (Uf - Ui) + (Kf - Ki) = 0 (0 - .5kd2) + (.5mv2 - 0) = 0 = .032 m 750 N/m .012 kg = 8.0 m/s

9. A 1.93 kg block is placed against a compressed spring on a frictionless 27.0˚ incline. The spring force constant is 20.8 N/cm and it is compressed 18.7 cm and released. How far up the incline will the block go before coming to rest? m = 1.93 kg K + Us + Ug = K0 + Us0 + Ug0 ø = 27.0˚ y = .5kx02 mg mgy = .5kx02 = 1.92 m k = 2080 N/m x0 = .187 m L = y sin Ø L = 4.23 m y

10. Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring loaded gun that is mounted on a table. The target box is 2.20 m horizontally from the edge of the table. Conor compresses the spring 1.10 cm, but the marble falls 27.0 cm short. How far should Mary Alice compress the spring to score a hit? let x = the spring compression ∆Us + ∆K = 0 let d = horizontal distance traveled .5k(x2 - xo2 ) + .5 m (v2 - vo2) = 0 kxo2 = mv2 = m(d /t)2 k,m,t are constants! d2 = kt2 x2 m d x is a constant ratio! so

11. Conservation of Energy in a System of Particles • When an object interacts with one or more objects we can choose our system to be as many or as few of those objects as is convenient. conservation of energy will always hold if care is taken to keep track of energies within the system and transfers in and out of the system! system system boundary U,K,Eint External W

12. The energy of the system within the boundary can be changed only when external work is done to it. (Internal work-- done by one part within the boundary acting upon another-- does not change the total energy of the system, although it may convert energy from one form to another, e.g. UK) Conservation of Energy can then be written: ∆U + ∆K +∆Eint = W

13. Consider a block attached to a spring sliding across a table that exerts frictional forces:

14. A) If the block itself is the system, then: ∆U + ∆K +∆Eint = Wtotal ∆K + ∆Eint = Ws + Wf B) If the block and the spring are the system: ∆Us + ∆K + ∆Eint = Wf C) If the block, spring and the table are the system: ∆U + ∆K + ∆Eint = 0 *the Conservation Laws hold in each case!

15. A baseball of mass .143 kg is dropped from a height of 443 m and reaches a terminal speed of 42 m/s. Find the change in internal energy of the ball and surrounding air during the fall back to the surface of the earth. If the system is regarded to be ball, the air and the earth, then no external forces or work are present: ∆U + ∆K + ∆Eint = 0 ∆Eint = - ∆U - ∆K

16. ∆U = Uf - Ui = mgh - mgho 0 - (.143)(9.8)(443) = - 621 J ∆K = Kf - Ki = .5mv2 - .5mvo2 .5(.143)(42)2 - 0 = 126 J Eint = - ∆U - ∆K - (- 621 J) - (126J) = 495 J The internal energy increase might be observed as a temperature increase of the air/ball, and/or an increase in the KE of the air the ball passes through.

17. A 4.5 kg block is thrust up a 30.0˚ incline with an initial speed of 5.0 m/s and it travels 1.5 m before coming to rest. A) How much mechanical energy does the block lose in this process due to friction?B) Assuming friction to produce the same loss in mechanical energy on the return downward journey, what is the speed of the block as it passes through its initial location? If we choose our system to be the block and the plane, then there are no external forces/work: ∆U + ∆K + ∆Eint = 0

18. In this case ∆U and ∆K would refer to the block, and therefore ∆Eint would refer to the energy converted to friction and thermal energy. ∆E = - ∆U - ∆K ∆U = mgh - mgho = (4.5)(9.8)(1.5 m)(sin 30) - 0 = 33 J ∆K = .5mv2 - .5mvo2 = 0 - .5(4.5)(5.0)2 = - 56 J Positive value would indicate an overall gain! ∆E = - (33 J) - (-56 J) = 23 J Therefore, the block lost 23 J of mech. energy!

19. Now if we consider the system to be just the block: ∆U + ∆K + ∆Eint = Wf The block would experience changes in U and K, and its loss of energy would be represented by: ∆U + ∆K = Wf - ∆Eint In this case, ∆U + ∆K = - 23 J, meaning the energy went out of the system (block). It would impossible to determine how much energy was lost due to Wf and how much was due to internal changes (thermal energy, etc).

20. B) Again consider the system to be the block, both going up and then coming back down the plane: ∆U + ∆K = Wf - ∆Eint If the block loses 23 J of energy going up (Wf - ∆Eint), then it will lose a total of 46 J going up and down: ∆U + ∆K = - 46 J Kf - Ki = - 46 J Kf = - 46 J + 56 J = 10 J vf = 2Kf m = 2.1 m/s

21. OR: from the top down ∆U = mg(h - ho) = 0 - (4.5)(-9.8)(-1.5 m)(sin 30) = - 33J ∆U + ∆K + ∆E = 0 ∆K = - ∆U - ∆E = - (-33 J) - 23 J = 10 J vf = 2Kf m = 2.1 m/s

22. A ski run connects to mountains separated by a valley. A skier starts from rest down the first hill from a height of 362 m and coasts until topping the second hill at a height of 131 m at which point he has a speed of 23 m/s. What is the change in internal energy of his skis and the snow over which he traversed?