{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Physics for Scientists and Engineers

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(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
Image of page 2
*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 .
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
*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. O CM 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
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern