**Unformatted text preview: **surface, UO+ Ko = U + K, or ~mV5 + mgyo = 4mv2 + mgy. (We neglect the effects of air resistance.) Here, the y-axis is posiiive upward, so that the gravitational potential energy, relative to zero potential energy at y = 0, has the usual form mgy. Then the distance dropped is Yo - Y = h, instead of x - Xo =h, and ay = -g. Canceling m, and rearranging terms, one recaptures the stated result. Problem
26. In a switchyard, freight cars start rrom:rest and roll down a 2.8-m incline and come to ~est against a spring bumper at the end of the track (Fig. 8-27). If the spring constant is 4.3X106 N/m, how much is the spring compressed wh~n hit by a 57,OOO-kgreight car? f Ii FIGURE &-28Problem 27. FIGURE 8-27 Problem 26. of the kids plus toboggan (including potential energy of gravitation. and the spring, as well as kinetic energy) is conserved: Utop + Ktop = Ubot + Kbot. At the top of the hill, Ktop= 0 (the toboggan starts from rest), and Utop = Utop,g + Utop,s = mgYtop + !kx2, while at the bottom, Kbot = ~mv2 and Ub...

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