phys for scient & eng serway 5th edition chpts 8-2

Physics for Scientists and Engineers

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8.38 (a) The mass moves down distance 1.20 m + x . Choose y = 0 at its lower point. K i + U gi + U si + E = K f + U gf + U sf 0 + mgy i + 0 + 0 = 0 + 0 + 1 2 kx 2 (1.50 kg)9.80 m/s 2 (1.20 m + x ) = 1 2 (320 N/m) x 2 0 = (160 N/m) x 2 – (14.7 N) x – 17.6 J x = 14.7 N ± (–14.7 N) 2 – 4(160 N/m)(–17.6 N · m) 2(160 N/m) x = 14.7 N ± 107 N 320 N/m The negative root tells how high the mass will rebound if it is instantly glued to the spring. We want x = 0.381 m
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(b) From the same equation, (1.50 kg)1.63 m/s 2 (1.20 m + x ) = 1 2 (320 N/m) x 2 0 = 160 x 2 – 2.44 x – 2.93 The positive root is x = 0.143 m (c) The full work-energy theorem has one more term: mgy i + fy i cos 180° = 1 2 kx 2 (1.50 kg) 9.80 m/s 2 (1.20 m + x ) – 0.700 N(1.20 m + x ) = 1 2 (320 N/m) x 2 17.6 J + 14.7 N x – 0.840 J – 0.700 N x = 160 N/m x 2 160 x 2 – 14.0 x – 16.8 = 0 x = 14.0 ± (14.0) 2 – 4(160)(–16.8) 320 x = 0.371 m 8.39 Choose U g = 0 at the level of the horizontal surface. Then E = ( K f K i ) + ( U gf U gi ) becomes: f 1 s f 2 x = (0 – 0) + (0 – mgh ) or –( μ k mg cos 30.0 ° ) ϒ 0 . 30 sin h – ( μ k mg ) x = mgh Thus, the distance the block slides across the horizontal surface before stopping is: x = h μ k h cot 30.0 ° = h ϒ 0 . 30 cot 1 k = (0.600 m) ϒ 0 . 30 cot 200 . 0 1 or x = 1.96 m h = 60.0 cm = 30.0 ° m = 3.00 kg θ
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*8.40 The total mechanical energy of the diver is E mech = K + U g = 1 2 mv 2 + mgh . Since the diver has constant speed, dE mech dt = mv dv dt + mg dh dt = 0 + mg (– v ) = – mgv The rate he is losing mechanical energy is then dE mech dt = mgv = (75.0 kg)(9.80 m/s 2 )(60.0 m/s) = 44.1 kW 8.41 U ( r ) = A r F r = – U r = – d dr A r = A r 2 8.42 F x = – x U ƒ ƒ = – x x y x ƒ ƒ(3 7 3 = –(9 x 2 y – 7) = 7 – 9 x 2 y F y = – y U ƒ ƒ = – y x y x ƒ ƒ(3 7 3 = –(3 x 3 – 0) = –3 x 3 Thus, the force acting at the point ( x , y ) is F = F x i + F y j = (7 – 9 x 2 y ) i – 3 x 3 j *8.43 (a) There is an equilibrium point wherever the graph of potential energy is horizontal: At r = 1.5 mm and 3.2 mm, the equilibrium is stable. At r = 2.3 mm, the equilibrium is unstable. A particle moving out toward r approaches neutral equilibrium. (b) The particle energy cannot be less than –5.6 J. The particle is bound if –5.6 J E < 1 J . (c) If the particle energy is –3 J, its potential energy must be less than or equal to –3 J. Thus, its position is limited to 0.6 mm r 3.6 mm . (d) K + U = E . Thus, K max = E U min = –3.0 J – (–5.6 J) = 2.6 J (e) Kinetic energy is a maximum when the potential energy is a minimum, at r = 1.5 mm . (f) –3 J + W = 1 J. Hence, the binding energy is W = 4 J .
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*8.44 stable unstable neutral 8.45 (a) F x is zero at points A, C and E; F x is positive at point B and negative at point D. (b) A and E are unstable, and C is stable. (c) x (m) F x A B C D E 8.46 (a) As the pipe is rotated, the CM rises, so this is stable equilibrium. (b) As the pipe is rotated, the CM moves horizontally, so this is neutral equilibrium. (c) As the pipe is rotated, the CM falls, so this is unstable equilibrium. OC M b O a CM O c CM 8.47 (a) When the mass moves distance x , the length of each spring changes from L to x 2 + L 2 , so each exerts force k ( x 2 + L 2 L ) toward its fixed end. The y - components cancel out and the x -components add to: F x = –2 k ( x 2 + L 2 L ) x x 2 + L 2 = –2 kx + 2 kLx x 2 + L 2
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phys for scient & eng serway 5th edition chpts 8-2 -...

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