**Unformatted text preview: **0.85)]j2(2.7) = 1.74 cm. (The negative root can be discarded since x ~ 0.) (b) The maximum kinetic energy occurs at the point where U is a niinhrtum; that is, dU I dx 2axo - b 0, or Xo = b/2a = O.778cm. (c) Vmax = J2Kmax/me = J2(E - Urnin)/me. Now, Umin = xo{axo - b} = = = = = = -b2/4a = -(4.2 fJ cm 24{2.7 fJ cm2) = -1.63 fJ, so that vmax = 2(0.85 + 1.63) fJ/{9.llxlO-31 kg) = 7.38x 107 m/s. u Problem
80. Show that the rope in Example 8-6 will cease to be taut when the rope has caught on the rock and makes an angle ()= cosl [~~ (cos 00 + ~ -19] Problem 81. Solution. with the vertical. Show that your anSM::f is consistent with that of Problem 33....

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