**Unformatted text preview: **ot = mgYbot (since Ubot.8 = 0 when the spring is no longer compressed). (b) Using the given data, we find Kbot = mg(YtopYbot) + ~kx2 = (80 kg)(9.8 m/s2)(9.5 m)+ !(890 N/m)(2.6 m)2 = 7.45 kJ + 3.01 kJ = 10.5 kJ, so Utop,s/Kbot = 3.01 kJjlO.5kJ = 28.8%. (a) The speed at the bottom is Vbot = y'2K bot/m = y'2(1O.5kJ)j(80 kg) = 16.2 m/s. Solution
The freight car starts and ends at rest (no kinetic energy),.so if we neglect friction, the work-energy theorem requires Wnet = t::..K = 0 = Wg +Ws' The work done by gravity is Wg = mg(2.8 m),andthe .work done by the spring force is Ws =-~ki2 (the spring force is opposite to the displacement). Therefore, x = .Y'2W;Jk = V2(5.7X104 kg)(9.8 m/s2)(2.8 m)/(4.3x106 NJm) = 85.3 em. (Of course, for conservative forces, + Us), and the work-energy theorem is just the conservation of mechanical energy.)
Wg + Ws = -tl(Ug Problem
28. A 20Q-gblock slides back and forth on a frictionless surface between two springs, as shown in Fig. 8-29. The left-hand spring has k = 130 N/m, and it...

View
Full Document