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253 m

10. 40.0N. 253 m. Only the component of force in the direction on the motion does work W = F d = cos52(40)(253) = 6230 J. Use force diagram to see that: F N + F  = W F N + sin52(40) = 70 F N = 38.5 N

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253 m

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  1. 10 40.0N 253 m Only the component of force in the direction on the motion does work W = F d = cos52(40)(253) = 6230 J Use force diagram to see that: FN + F = W FN + sin52(40) = 70 FN = 38.5 N Ff = FN cos52(40) =  (38.5)  = .640 If she is moving at constant speed, the friction force must equal F|| and the distance for friction is also 253 m, so the WORK must also be the same put in the opposite direction. -6230 J (I don’t get too worried if you don’t have the negative sign)

  2. 16 50 km/hr A B 50km 1000 m 1 hr hr 1km 3600 s 35 m --------x -----------x-----------= 13.9 m/s EA - Efriction = EB ½mv2 - Ffd = 0 ½m(13.9)2 - (9.8m)(35) = 0 ½m(13.9)2 = (9.8m)(35) .282 =  Same situation, but with double the speed: 27.8 m/s ……..find d ½m(27.8)2 = (.282)(9.8m)d 140 m = d

  3. 22 B m= 20 W=196 4 m/s h d? A 25 sin25=h/d EA - Efriction = EB KE - W = PE ½ m v2 - Ff d = mgh FN ½(20)(4)2 - .2(cos25)(196)d = 20(9.8)(sin25)d 160 - 35.5d = 82.8d d = 1.35 m

  4. 23 This problem is just proving that it doesn’t matter where you draw the “zero line” A PEA = 55(9.8)(10) = 5390 PEB = 0 PE = PEA - PEB = 5390J 10 m B Ref pt A B. PEA = 0 PEB = 55(9.8)(-10)=-5390 PE = PEA - PEB = 5390J Ref pt 10 m B C. PEA = 55(9.8)(5) = 2695 PEB = 55(9.8)(-5)=-2695 PE = PEA - PEB = 5390J A 10 m Ref pt B

  5. 33 A 7.34 v? B EA = EB PEgrav = KE mgh = ½ mv2 9.8(7.34) = .5v2 12.0 m/s = v

  6. 34 A. EA = EB 37 B. EA = EB

  7. 35

  8. 36

  9. 47 4.0m/s 70kg  = .70 d? EA - Efriction = EB KE - W = 0 ½ mv2 - Ffd = 0 ½ (70)(4)2 - .70(70x9.8)d = 0 560 - 480d = 0 d = 1.17 m ALL of the KE was lost due to friction, so the answer to part A is 560J

  10. 49

  11. 54 m = m W = 9.8m W|| = sin10.5(9.8m) = 1.79m W = cos10.5(9.8m) =9.64m EA - Efriction = EB PEgrav - Wf - Wf = 0 on hill on flat mgh - Ffd - Ffd = 0 m(9.8)(36.4) - .075(9.64m)(200) - .075(9.8m)d = 0 d = 288 m

  12. 59 B A 2 kg 2 kg .25 m  = ? k = 105 N/m x = .10 m EA – Efriction = EB PEspring – Wfriction = 0 ½ kx2 - Ff d = 0 ½ (105)(.1)2 - (29.8)(.25) = 0 4.9 = .525  = .107

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