2015 USA Physics Olympiad Exam Part A 8 Take the derivative,and △E=kA△A That energy is lost to friction.if AAc then the top block has almost completed a complete half cycle before the bottom block catches up with it (speed),so the energy lost in half a cycle is approximately 1 △E=2Af=2Aμmtg 2 where f is the frictional force. Combine, 4μmtg=k△A, or △A=4m9=49 42 c.Assume still that A>Ac.What is the maximum speed of the bottom block during the first complete oscillation cycle of the upper block? Solution The bottom block accelerates according to mt a=μ mb The upper block oscillates as if it is free,since the bottom block exerts a constant force on it,so wt=Vk/mt gives a half period of t=TVmt/k. The maximum speed is then mt b=μTVmt/R mh Note that the maximum speed of the top block is Ut Awt The ratio is =μ9mbkA mtmt t Remember that Ae=μ9K t mt 1+ b %=Am A mb+mt ←1 Copyright C2015 American Association of Physics Teachers2015 USA Physics Olympiad Exam Part A 8 Take the derivative, and ∆E = kA∆A. That energy is lost to friction. if A  Ac then the top block has almost completed a complete half cycle before the bottom block catches up with it (speed), so the energy lost in half a cycle is approximately 1 2 ∆E = 2Af = 2Aµmtg where f is the frictional force. Combine, 4µmtg = k∆A, or ∆A = 4µmtg k = 4 µg ωt 2 c. Assume still that A  Ac. What is the maximum speed of the bottom block during the first complete oscillation cycle of the upper block? Solution The bottom block accelerates according to a = µg mt mb The upper block oscillates as if it is free, since the bottom block exerts a constant force on it, so ωt = p k/mt gives a half period of t = π p mt/k. The maximum speed is then vb = µg mt mb π p mt/k Note that the maximum speed of the top block is vt = Aωt The ratio is vb vt = µg mt mb mt kA Remember that Ac = µg mt k  1 + mt mb  , so vb vt = Ac A mt mb + mt  1 Copyright c 2015 American Association of Physics Teachers