**Unformatted text preview: **entialenergy difference t::"U == U B UA = -mg(YA - VB), where YA -VB = 7.2 m. The wark dane by frictian( on level ground, N == mg) is Wnc = - fkX = -/lkmgx, where x is the distance slid across the rough level stretch. The energy principle; Equatian 8~5, relates these quantities: Wnc -/lkmgx = b.K + l:i.U =0 ~ mg(YA - YB).Thus x = (YA - YB)//lk =7.2 in/0.51= 14.1 far X2 < Xl. If the kinetic energy af an alpha particle at the nuclear surface is zero, K2 0, then its kinetic energy at same other paint is K1 = U2 - UI. Therefare, its speed is = VI = 1 2K ! = 2A
mX2 m (1 X2)
Xl 2(4.1xlO-26N.m2) = [ (6.7xlO-27 kg)(1.43xlO-10 = (2.93><105 .( m) 1- X2)]1/2
Xl rn/s)V1- X2/XI, = m. where X2 is the nuclear radius. (a) If Xl = 4x2, VI = 5 2.53x10 mfs, and (b) if Xl = lOOX2, VI = 2.91 x 5 10 m/s. (c) Evidently, 2.93x105 m/s= VI for
Xl -+ 00. Problem
74. At the bottom of a frictianless ski slope is a 20-m-wide stretch af rough snow where the caefficient af frictian is 0.32. Fram what vertical...

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