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Unformatted text preview: P7.6 (a) Energy of the particleEarth 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)(200) = FIG. P7.6 (b) AKP_,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 blockspring—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‘ nghfh=§m1vz +Zmzv f=ﬂn =#m1g 1
nghmlgh=—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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 Winter '07
 MahaAshourAbdalla

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