8 ilium height of 5 m 136 m ms 136 m to a 2

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Unformatted text preview: he floor, e a non-conservative contact force acts to stop ;Tlle kinetic energy of the 7-kg mass is lost strikes the floor. !(7 kg)(5.17m/s)2 93.5 J 111;27.3% of the initial energy of the system, i.;,yAs (7x9.8 N)(5 m) = 343 J. -it = FIGURE 8-34 Problem 38. ;~11l ';~ass m is dropped from a height h above the :~f.a spring of constant k that is mounted ;lc&ilyon the floor (Fig. 8-33). Show that the 'mum compression of the spring is given by > Jk)(l + + 2kh/mg). What is the ~i1icance of the other root of the quadratic uation? VI the reference level y = 0 in Fig. 8-34), so in the absence of friction, the sum of these is a constant. We are given that VA = 0 and YA = 3.8 m, so we can evaluate the constant and express the mechanical energy at any other point in terms of it: ~mv2 + mgy = mgYA. (a) Solving for the speed at point H, we find VB = V29(YA - YB) = m - 2.6 m) = 4.85 m/s. (b) Similarly, Vc = 7.00 m/s. (c) The right-hand turning point is the point D where the particle's velocity is instantaneously zero, before changing dire...
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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.

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