Work, Energy, Power, and Simple Machines

1. Work, Energy,Power, andSimple Machines Work …done on an object when a force moves object through a distance d = distance object moved (m) W = F d F = applied force // to d (N) Unit for work is N-m, or J. F and dmust be //.

2. Fa, d Fw = m g Fw = m g Student carries 63.4 g tissue box 3.2 m at constant height of 1.08 m above floor. How much work does gravity do on box? d ZERO; Fw and d are A 63.4 g facial tissue box is lifted at constant speed from the ground to height 1.08 m. Find work the lifter does on box. W = Fa d = m g d = 0.0634 kg (9.81 m/s2) (1.08 m) = 0.672 J

3. = 170 N (3.4 m) Fa, d Fa mk Fn m g Fn 25 kg mass is pulled 3.4 m across horiz. surface by 170 N horiz. force. Coefficient of friction is 0.33. Find work done by applied force. W = Fa d m = +580 J mk Find work done by friction. Draw FBD… m = –280 J

4. Fa m mk Fn m g Fn Find work done by gravity. 0 J Find net work done on mass. 300 J Wnet = + 580 J – 280 J = Or… Wnet = Fnet d = (170 N – 81 N)(3.4 m) = 300 J

5. d 17.5 kg mass is pulled 5.43 m by a 185 N force applied at 21.0o above horizontal. If coefficient of friction is 0.420, find work done by every force acting on mass. Q Fa Fa sin Q Fa cos Q mkFn Fn mg m Wk = Fkd = 44.3 N (5.43 m) mk Draw FBD… Fn and Fw do no work. (66.3 N) comp. of Fa also does no work. m (172.7 N) (44.3 N) Work done by... ...friction (105.4 N) (171.7 N) = –241 J

6. ...applied force Wa = 172.7 N (5.43 m) = 938 J Find net work done on mass. Wnet = 938 J – 241 J = 697 J = 697 J Wnet = Fnet d = (172.7 N – 44.3 N)(5.43 m)

7. Energy the ability to do work (joules) -- unit is J 1 J is equivalent to 1 N-m. kinetic energy: energy of motion -- KE = ½ m v2 m = mass (kg) v = speed (m/s)

8. elastic potential energy (PEelas) gravitational potential energy (PEg) -- -- Dx stored energy potential energy: due to object’s location in a gravitational field internal energy due to “springy” deformation PEg = m g h PEelas = ½ k (Dx)2 h = height of object above arbitrary reference line k = spring constant (N/m) Dx = change in length (m)

9. Law of Conservation of Energy Energy remains constant, but it may change form. We can use this idea in solving problems if there is… …little or no friction (including air resistance) …no collision in which significant heat is released Procedure: 1. Take two snapshots of system. 2. Calculate various energies (KE, PEg, PEelas) for each snapshot. 3. The sum of these (i.e., the total mechanical energy, ME) for one snapshot equals the sum of these for the other snapshot.

10. m A B KE PEg PEelas ME vi 3.0 kg mass slides off a 1.8 m-high shelf at 6.0 m/s . Find mass’s speed as it hits ground. ½ m vi2 “A” h vf = ? “B” R.L. ½ m vf2 m g h 0 0 0 ½ m vf2 ½ m vi2 + m g h = ½ vi2 + g h = ½ vf2 vi2 + 2 g h = vf2 vf = 8.4 m/s

11. 3.0 kg mass slides off a 1.8 m-high shelf at 6.0 m/s . Find mass’s speed as it hits ground. A B KE PEg PEelas ME 3.0 kg ½ (3) (6)2 6.0 m/s “A” 1.8 m vf = ? “B” R.L. ½ (3) vf2 3 (9.81) (1.8) 0 0 0 1.5 vf2 107.0 = 71.3 = vf2 vf = 8.4 m/s

12. “B” h1 m A B KE PEg PEelas ME A 7.4 kg cat sits on window ledge 1.1 m above the floor on 13th story of apt. bldg. Floor level is 41.0 m above street. Find vel. of cat as it lands on floor. Cat starts from rest. vf = ? R.L. , vi h2 “A” 0 ½ m vf2 m g h1 0 0 0 = m g h1 ½ m vf2 g h1 = ½ vf2 = 4.6 m/s

13. h1 m A B KE PEg PEelas ME “B” A 7.4 kg cat sits on window ledge 1.1 m above the floor on 13th story of apt. bldg. Floor level is 41.0 m above street. Find speed of cat as it lands on floor. Cat starts from rest. vf = ? R.L. , vi h2 “A” 0 ½ (7.4) vf2 0 7.4 (9.81) (1.1) 0 0 = 3.7 vf2 79.85 21.58 = vf2 vf = 4.6 m/s

14. 3.0 m 1.5 m A B KE PEg PEelas ME A 2.0 m pendulum hangs from ceiling that is 3.0 m above floor. If 8.0 kg bob is released from rest 1.5 m above floor, find its speed at bottom of its path. 2.0 m “A” vf = ? 8.0 kg vi = 0 “B” R.L. ½ (8) vf2 0 8 (9.81) (1) 8 (9.81) (1.5) 0 0 4 vf2 + 78.48 117.72 = 39.24 = 4 vf2 9.81 = vf2 vf = 3.1 m/s

15. 6000 N/m h = ? A B KE PEg PEelas ME “B” A 0.38 kg object is at rest on a spring w/spring constant 6.0 x 103 N/m. Original length of spring is 82 cm, but it has been compressed to 52 cm. When spring is released, to what height will object go? Dx = 0.30 m 0.38 kg vi = 0 “A” R.L. 0 0 0.38 (9.81) h 0 ½ (6000) (0.3)2 0 270 3.73 h = h = 72 m

16. A B KE PEg PEelas ME Horiz. spring is attached to wall. An 8.5 kg mass is slid across a frictionless, horiz. surface at 2.8 m/s. When mass hits spring, spring compresses 45 cm. If surface is 1.4 m above ground, find spring constant. k = ? 2.8 m/s 0.45 m 8.5 kg R.L. R.L. frictionless 1.4 m “A” “B” 0 ½ (8.5) (2.8)2 0 0 0 ½ k (0.45)2 = 0.10125 k 33.32

17. 35 m/s 45 m/s Q 35 m/s “B” h = ? Arrow of unknown mass is launched at an unknown angle w/initial velocity 45 m/s. At top of its path, its speed is 35 m/s. How high does arrow fly? “A” R.L. 45 m/s m (9.81) h = + ½ m (45)2 ½ m (35)2 = 612.5 + 9.81 h 1012.5 400 = 9.81 h h = 41 m At what angle was arrow launched?

18. The Work-Kinetic Energy Theorem When friction and other heat losses are NOT negligible, the conservation of energy procedure is invalid. Thus, we turn to the work- kinetic energy theorem (which could, incidentally, be used in any of the previous examples). Wnet = DKE Fnet d = ½ m (vf2 – vi2)

19. 72 0 52 238 N From rest, a 72 kg object is pulled on a level surface by a horiz. 238 N force. If coeff. of kinetic friction is 0.28, how far will object have traveled by the time its speed reaches 5.0 m/s? Fnet d = ½ m (vf2 – vi2) mk = 0.28 72 kg 197.77 N d = 22 m 706.32 N

20. A 12 kg object slides on ice w/initial speed 2.2 m/s. If coeff. of kinetic friction is 0.030, how far will object slide? 2.2 m/s 2.22 12 d = 8.2 m Fk and d are in opposite senses! mk = 0.030 Fnet d = ½ m (vf2 – vi2) Draw FBD… – 3.53 d = ½ (12) (– 2.22) 12 kg 0 3.53 117.72

21. 4.0 m 0 12.5 kg A 12.5 kg mass begins from rest at top of a frictionless, 4.0 m-long, 32o incline. Find speed at bottom. vi = 0 h mk = 0 vf = ? 32o R.L. ½ (12.5) vf2 12.5(9.81)(4 sin 32) = 259.93 = 6.25 vf2  vf= 6.4 m/s Now find vf when mk = 0.27. Fnet d = ½ m (vf2 – vi2) 104 N 28.08 N (4) 36.92 = ½ (12.5) (vf2) 104 N vf = 4.9 m/s 65 N

22. = 7900 s Power  the rate at which work is done Unit for power is the Watt (W). A flooded basement contains 1.2 x 108 kg of water that must be pumped upwards 2.5 m. In what time would a 373 kW pump empty the basement? = 2.2 h

23. tA tB d m An 85 kg mass must be raised to height 4.5 m. Person A does so in 1 minute, 24 seconds; Person B does so in 2 minutes, 48 seconds. Find work AND power done by A AND by B. = 45 W WA = FAd = m g d = 85 kg (9.81 m/s2) (4.5 m) = 3.8 x 103 J WB = 3.8 x 103 J = 22 W

24. …work done by B. WB = F d = m g d = 85 kg (9.81 m/s2) (4.5 m) = 3.8 x 103 J …power expended by B. = 23 W

25. 1500 N 1500 N Feng 4905 N A 5.0 x 102 kg rocket accelerates upward from rest to 540 m/s in 2 minutes, 46 seconds. If air resistance is 1.5 x 103 N, find avg. power of rocket engine. Dt = 166 s 500 kg vi = 0 Peng = ? vf = 540 m/s Fnet = m a = 1626.5 N Fnet = Feng – 1500 – 4905 1626.5 So…Feng = 8031.5 N Peng = Feng vavg = 8031.5 (270) Peng = 2.2 x 106 W = 2.2 MW

26. Wnet = DKE W = F d Fw= m g Fnet d = ½ m (vf2 – vi2) KE = ½ m v2 PEg = m g h PEelas = ½ k (Dx)2