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# hw6s - P7.6(a Energy of the particle-Earth system is...

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Unformatted text preview: P7.6 (a) Energy of the particle-Earth system is conserved as the particle moves between point P and the apex of the trajectory. Since the horizontal component of velocity is constant, 1 1 1 Emviz = —mvix2 + Emviyz = émvix2 + mgh my = 2(9.80)(20-0) =- FIG. P7.6 (b) AK|P_,B = Wg = mg(60.0 m) = (0.500 kg)(9.80 m/sz)(60.0 m) =- (c) Now let the final point be point B. vxi = vxf = 30.0 m/s 2 :2ny = ;(294) + 2;in = 1 176 + 392 vyf = ‘39.6 III/S ‘73 = (30.0 m/s)i—(39.6 m/s)i P7.9 Using conservation of energy for the system of the Earth and the two objects (a) (5.00 kg)g(4.00 m) = (3.00 kg) g(4.00 m)+%(5.00 +3.00)v2 (b) Now we apply conservation of energy for the system of the 3.00 kg object and the Earth during the time interval between the instant when the string goes slack and the instant at which FIG. 137.9 the 3.00 kg object reaches its highest position in its free fall. %(3.00)v7‘ = mg Ay = 3.00gAy Ay = 1.00 m ymax =4.00 m+Ay=- P7.16 P7.17 P7.23 P7.25 Choose the zero point of gravitational potential energy of the obj ect—spring—Earth system as the conﬁguration in which the object comes to rest. Then because the incline is frictionless, we have EB =EA: KB +UgB +1153 =KA +UgA +LISA or 0+mg(d+x)sin0+0=0+0+—:—kxz. Solvingfordgives kxz d: —_—x . 2mgsm0 From conservation of energy for the block-spring—Earth system, 1131‘ = ”51" or (0.250 kg)(9.80 m/s2)h = (3(5 000 N/m)(0.100 m)2 This gives a maximum height h = . 1 1 2 Hi +Ki +AEmech =uf +Kf‘ ngh-fh=§m1vz +-Z-mzv f=ﬂn =#m1g 1 ngh-mlgh=—2—(m1 +mz)‘uz U2 zw FIG. P7.23 m1 +7712 2 . .00 k —0.400 3.00 k v: 2(9.80 m/s )(150 m)[5 g ( g)] = 3.74 m/S 8.00 kg (a) AK=%m(v%—vi2)=-—12—mvi2=- (b) All = mg(3.00 m) sin30.0°= 73.5 J (c) The mechanical energy converted due to friction is 86.5] = 86.5 J = FIG. P7.25 (d) f = mm = ,ukmgc0530.0°= 28.8 N 28.8 N = ———-——— = ".679 ”k (5.00 kg)(9.80 m/sz)cos 300° P7.51 AEmech = —fd ’1 = =- FIG. P7.51 ...
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hw6s - P7.6(a Energy of the particle-Earth system is...

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