phys141 s06 white exam

# phys141 s06 white exam - Answers(WHITE exam 1(10 points...

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Unformatted text preview: \ Answers (WHITE exam) 1] (10 points) Mottmann weighs a muscular 200 lbs. To lift him up a height H requires energy = mgH = (200)H ft-lbs. There is 600 Cal of energy in the burger = (600x3100) = 1.9x106 ft-lbs, or H = 1.9x106/200 = 9300 ft = 1.8 miles. 2] (5 pts each) 3) always less than Mg {N = Mgcose} b) always greater than Mg {N = Mg/sine; work it out!} V 3] (10 pts)See Worked Example 121 use conservation of energy: Bend = Estart or 1/2mc2 — GmM/RS = 0 or Rs = 2GM/c2 correct even in General Relativity! 4] (5 pts each) a) B {B travels faster on every part of the “down” bump compared to A on the “up” bump} b) less than 1 m/s {in fact, A can’t make it over the top with the numbers given; work it out!} 5] a) (10 pts) see diagram at right N1/2 = normal force between carts ,r N 1/earth = normal force between cart I and earth ,’ N2/earth = normal force between cart 2 and earth l‘ ‘1. Push = force between ﬁnger and cart 2. M1 g and Mg are the forces of gravity @i’ 0 pts) consider only cart 1 (see the dashed line in the ﬁgure) \V’ fo=ma 01' NI/2=M1A that’s all there is to it, we’re done! 6] a) (10 pts) There are two objects: MA and MB. I can pick a different coordinate system for each, as is shown in the ﬁgure at right: Each block starts at its own height: 0 and is not moving, so Estart = 0. V Each block moves a distance (up or down) of 2 m and the spring compresses a distance of l m, so Eend = 1/2k(1)2 + M Ag(2) - MB g(2). ﬁnally: Eend = Estart or k = 39.2 N/m b) (10 pts) The blocks are shown at right at the instant they are at rest Gravity pulls and the spring pushes: forA: Efy = ma or T - MAg= - MAa for B: ny = ma or T - MBg + kAy = MBa where Ay is the compression of the spring solve the two equations for a: aIMA + M3] = g(MA — MB) + My or ﬁnally a = 5.9 m/s2 @ (10 pts) Bend = Estart or l/2Mv2 = Mgh or I/2Mv2 = Mg(2/3R) or v2 = 4/3gR or v = 1.6 m/s b) (10 pts) Bend = Estart - wfric or Mgh = Mgho - uNAx or h = ho - (uNAx)/Mg Where h0_= 2/3R = 0.13 m and Ax = 1 cm = 0.01 m to ﬁnd the normal when the block is at the bottom: ZfRAD = maRAD or N - Mg = MV2/R or N = M(g + v2/R) = 7/3Mg ﬁnally: h = h0 — (mm/Mg = ho- (”7/3MgAx)/Mg = ho - 7/3qu = h0 - 0.01 m = 0.13 — 0.01 = 0.12 m ...
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