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253 m A. Only the component of force in the direction on the motion does work W = Fd = cos52(40)(253) = 6230 J B. 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) C. Use force diagram to see that: FN + F = W FN + sin52(40) = 70 FN = 38.5 N Ff = FN cos52(40) = (38.5) = .640 A 50 km/hr B 50km 1000 m 1 hr --------x -----------x-----------= 13.9 m/s hr 1km 3600 s 35 m EA - Efriction = EB ½mv2 - Ffd = ½m(13.9)2 - (9.8m)(35) ½m(13.9)2 0 = 0 = (9.8m)(35) .282 = Same situation, but with double the speed: ½m(27.8)2 27.8 m/s ……..find d = (.282)(9.8m)d 140 m = d m= 20 W=196 sin25=h/d 25 EA KE - ½ m v2 - ½(20)(4)2 - 160 - Efriction = W = Ff d FN = EB PE mgh .2(cos25)(196)d = 20(9.8)(sin25)d 35.5d = d = 82.8d 1.35 m A This problem is just proving that it doesn’t matter where you draw the “zero line” PEB = 0 10 m B Ref pt A A. PEA = 55(9.8)(10) = 5390 Ref pt PE = PEA - PEB = 5390J B. PEA = 0 PEB = 55(9.8)(-10)=-5390 10 m B PE = PEA - PEB = 5390J C. PEA = 55(9.8)(5) = 2695 A 10 m B Ref pt PEB = 55(9.8)(-5)=-2695 PE = PEA - PEB = 5390J 7.34 v? EA = EB PEgrav = KE mgh = ½ mv2 9.8(7.34) = .5v2 12.0 m/s = v 37 A. EA = EB B. EA = EB 70kg 4.0m/s = .70 EA KE 2 ½ mv ½ (70)(4)2 560 d? - Efriction = EB W Ffd = = 0 0 = 0 - .70(70x9.8)d - 480d = 0 d = 1.17 m ALL of the KE was lost due to friction, so the answer to part A is 560J m = m W = 9.8m W|| = sin10.5(9.8m) = 1.79m W = cos10.5(9.8m) =9.64m EA - PEgrav - mgh Efriction Wf - Ffd - on hill - Wf = EB = 0 on flat Ffd = 0 m(9.8)(36.4) - .075(9.64m)(200) - .075(9.8m)d = 0 d = 288 m 2 kg k = 105 N/m x = .10 m EA – 2 kg .25 m =? Efriction = EB Wfriction Ff d = = 0 0 ½ (105)(.1)2 - (29.8)(.25) = 0 PEspring ½ kx2 – - 4.9 = .525 = .107