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# ch6 - P6.1(a w FA cos.9 ={16.0 N{2.2n m)cos 25.00 = 31.9...

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Unformatted text preview: P6.1 (a) w : FA:- cos .9 = {16.0 N){2.2n m)cos 25.00 = 31.9 J (b), (c) The normal force and the weight are both at 90“ to the displacement in any time interval. Both do [El work (d) ZW:31.9]+U+O: 31.9} P63 Method One. Let 92\$ represent the instantaneous angle the rope makes with the vertical as it is swinging up from \$5,- : 0 to \$5}. 260° . In an incremental bit of motion from angle (:5 to it + 11425 , the deﬁnition of radian measure implies that Ar : (12 m)d¢. The angle Qbetween the incremental displacement and the force of gravity is 9 = 90°+ 415. Then cos 9 = cos[90°+¢) = — Sinai . The work done by the gravitational force on Batman is FIG. P63. f 260° w = JFcosEHr = ﬂ [mg{—sm¢)(12 m)d¢5 i 19:0 : —mg(12 m) jsina as = [—30 kg)[9.s m/sz)(12 m)[— cos a) D = (—734 N)(12 m)(— cos6ﬂ°+1) = 600 D Method Two. The force of gravity on Batman is mg : [80 kg)[9.8 111/52) : 784 N down. Only his vertical displacement contributes to the work gravity does. His original y—coordinate below the tree limb is —12 In. His ﬁnal y—coordinate is [—12 m) cos 60" = —6 In. His change in elevation is —6 m— (—12 m) : 6 m . The work done by gravity is w = amass = (784 N)(6 m) cos 130°: 41.70 k] . P6.11 W = jdex and W equals the area under the Force—[ﬁsplaeement curve 3 (a) (b) (C) (d) P1531 (a) (b) (C) (d) (B) For the region 0 5x 1: 5.00 m , (m) _ (3 .00 N)(5.00 m) _ 0 5 10 15’ W — f — 7.50 I For the region 5.00 S x 510.0 , “(3' P631 W : (3.00 N)[5.00 m) : 15.0 I For the region 10.0 S x S 15.0 , _ [3.00 N)(5.00 m) _ W — f — 7.50 I For the region 0 ﬁx 5 15.0 W : [750+ 7.50 +150) I: 30.0} (0) KA = \$0.000 kg}[2.00 IIy’Sf :- 1 2 {b} Era-ms IKE: 'LJ' [2K3 {(200.50} B: m 2 0.000 :- (c) zwzaKzKB—KA=—m[u§—0ﬁ)=?.50]—1.20}=- P632 2 Fy : may: :1 + (70.0 N) sin 20.0°—147 N : 0 n. = 123 N fk = #0: = 0.300(123 N): 36.9 N w = FArcosQ: (70.0 N)(5.l] m)cos20.0°= 329 I w : FArcosQ: (123 N)(5.00 m)cos90.0°: W = FAr £059: (147 N)(5.00 m)cos90.0°= mam : fkd : (36.9 N)(5.00 m) = 105 I AKsz—Kj :ZW—AE- :329]—‘185] 0' FIG. P632 =- P636 (a) Z W = AK , but AK = 0 because he moves at constant speed. The skier rises a vertical distance of (60.0 m} sin 300°: 30.0 m. Thus, W... = 4w: = (70.0 kg)[9.8 m/sz)[30.0 m): 2.0mm“ I = 20.6 k] . (b) The time to travel 60.0 m at a constant speed 0152.00 mfs is 30.0 5. Thus, w 2.06x10‘] “gainful ZEZWZ-Zﬂglg hp. ...
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