{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online