This preview shows page 1. Sign up to view the full content.
Unformatted text preview: surface, UO+ Ko = U + K, or ~mV5 + mgyo = 4mv2 + mgy. (We neglect the effects of air resistance.) Here, the yaxis 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 2.8m incline and come to ~est against a spring bumper at the end of the track (Fig. 827). If the spring constant is 4.3X106 N/m, how much is the spring compressed wh~n hit by a 57,OOOkgreight car? f Ii FIGURE &28Problem 27. FIGURE 827 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...
View
Full
Document
This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.
 Spring '08
 WORMER
 Physics, Conservation Of Energy, Energy, Work

Click to edit the document details