B show that as long as the string remains taut the

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Unformatted text preview: eaving the ramp (at point 2), with speedv2 at 45 to the horiZontal, the block describes projectile motion with a horizontal range of x = vUg (see Equation 4-10). Since the track is frictionless (and the normal force does no work), the mechanical energy of the block (kinetic plus gravitational potential) is conserved between point 2 and its start from rest at point 1. Then !)'K = 0 = -!).U = mg(h1 - h2), and x = 2(h1 ~ h2). 0 Solution (a) At the top of the circle, the forces acting on the mass are gravity and the string tension, both downward and parallel to the centripetal acceleration. Thus T + mg = mv;op/ R. Since nop ~ 0 if the string is taut, v;op ~ gR. (See Example6-8.) (b) The mechanical energy of the mass is conserved, since the tension does no work (by assumption), gravity is conservative, and air resistance is ignored. Thus Utop + Ktop = Ubot + Kbot ~mv~op + mgytop = !mv~ot + mgYbot, Or V~ot = v;op + 2g(Ytop - Ybot)= !mv~- = FIGURE 8-42 Problem 65 Solution. Problem 66. A block of mass m is launched horizontally from a compres...
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