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# hw5s - P6.1(a w = FAr c050(16.0 N(2.20 m)cosZS.O°= 31.9...

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Unformatted text preview: P6.1 (a) w = FAr c050: (16.0 N)(2.20 m)cosZS.O°= 31.9 I (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]+0+0=- P6.12 Compare an initial picture of the rolling car with a ﬁnal picture with both springs compressed K + 2W = K f . Work by both springs changes the car’s kinetic energy Ki +%k1(x1,~2 "7‘11‘2)+%k2(x2i2 'foz) =Kf é—mviz + 0 — %(1600 N/m)(0.500 m)2 l 2 é“ 000 kg)vi2 —200 I—68.0]=0 ’ 2(268 I) . = = 0.299 vi 6 000 kg + 0 — (3400 N/m)(0.200 m)2 = 0 FIG. P6.12 P6.13 k=5=M§=W=1.57x103N/m y 1 2.50 x 10-2 111 mg (1.50)(9.80) F 1.50k =—=—-—= 0.938 cm (a) or g mass y k 1.57 )< 103 (b) Work = ikyz Work =—:—(1.57x103 N-m)(4.00 x10’2 m)2 = 1.25] P6.27 2w =AK =0: JOngsinSSD° dl— jjkx dx=0 mg sin 35.0° (L) = %kd2 d_ [2mgsin35.0°(L) k [202.0 kg)(9.80 m/s.2 )(sin35.0°)(3.00 m) d = _______ = 0.116 3.00x104 N/m P6.30 P632 P655 1 (a) Ws=§ kxi——kxf=—(500)(5.00x10'2)2 —0= 0.625] k=500N,m ([ (W Ws=lm vz_lmvz=1mvf_0 unwr- 2 2 17:2} (2W) =.){2(0625 2.00 m/s= _.791m/s 1 (b) —2—mv 25—fkd+W =§mvf 0 — (0.350)(2.00)(9.80)(0.050 0) 1+ 0.625 J: imv? 0.282 I: am) 1%)ng FIG. P6.30 _ 2(0.282) 71f - 2‘00 III/S: 0.531 m/s 3 O ‘ 00 ZFy = my: n+(70.0 N)sm20.0 —147 N =0 ﬁ (72)} N) sm 2 -—-———-:> (70 N) cos 20° 11 = 123 N fk = ,ukn = 0.300(123 N) = 36.9 N (a) w = FArc050= (70.0 N)(5.00 m)cos20.0°= 329 J (b) w = FAr cos 9: (123 N)(5.00 m) cos 900°: ‘m§=147N Ar = 5.00 m———> FIG. P6.32 (c) w = FAr cos a = (147 N)(5.00 m) cos 900° = [E] (d) AEint = fkd = (36.9 N)(5.00 m) =- (E) AK=Kf—Ki=ZW'AEint=329]_185]=- Ki +WS +Wg =Kf gmviz +§kxi2 —%kx§? +mgAxcos¢9=§mvj§ 0 + % kxiz — 0 + mgxi cos 100° = émv? FIG- P655 %(1.20 N/cm)(5.00 cm)(0.050 0 m) — (0.100 kg)(9.80 m/sz)(0.050 0 m) sin 10.0°= \$0.100 1<g)v2 0.150 ]— 8.51 x 10‘3 I= (0.050 0 1<g)v2 0141 0.050 0 = z): ...
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hw5s - P6.1(a w = FAr c050(16.0 N(2.20 m)cosZS.O°= 31.9...

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