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Unformatted text preview: 6-34[Leg Presses](a) First, find the spring constant k:Wby you= - Wby spring= 12kx2- 0 ⇒k = 2Wby springx2= 2(80.0 J)(0.200 m)2= 4000 N/mNow, the force: F = kx = (4000 N/m)(0.200 m) = 800 N(b) Wby you= - Wby spring= 12kxf2- 12kxo2= 12k(xf2- xo2)= 12(4000 N/m) [(0.400 m)2- (0.200 m)2] = 240 J7-12[Tarzan and Jane]Take y = 0 to be at the bottom of Tarzan's swing. Hestarts at height yo= L - L cos45° and descends to finalheight yf= L - L cos30°.Use Cons. of Energy to find Tarzan's final speed:Ko+ Ugo= Kf+ Ugf0 + mgyo= 12mvf2+ mgyfSolve for vf= 2g(yo- yf) = 2gL(cos30° - cos45°)= 2(9.8 m/s2)(20.0 m)(0.866 - 0.707) = 7.9 m/sThis is pretty fast. Tarzan knocks Jane off her feet like a football player making a tackle. Ahh-ee-aahhh!!!7-24[Satellite Launcher](a)The maximum acceleration occurs when the net force on the satellite is maximum. This iswhen the spring is fully compressed.Applying N’s 2ndLaw:Fmax= k xmax= m amax(1)When the satellite is going at its maximum speed, all of the PE initially stored in the spring has beenconverted to KE of the satellite:_ k xmax2= _ m vmax2(2)Using Equs (1) and (2), we can find both k and xmax....
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