{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Fifth HW Solution - CHAPTER 13 Problem Solutions 13.1(a The...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
C HAPTER 13 Problem Solutions 13.1 (a) The force exerted on the block by the spring is ( 29 ( 29 160  N m 0.15 m 24 N s F kx = - =- - = + , or s F = 24 N directed toward equilibrium position (b) From Newton's second law, the acceleration is 2 24 N m 60  0.40 kg s s F a m + = = = + = 2 m 60   toward equilibrium position s 13.7 (a) The spring constant of each band is 3 -2 15 N 1.5 10   N m 1.0 10  m s F k x = = = × × Thus, when both bands are stretched 0.20 m, the total elastic potential energy is ( 29 ( 29 2 2 3 1 2 1.5 10   N m 0.20 m 2 s PE kx = = × = 60 J (b) Conservation of mechanical energy gives  ( 29 ( 29 s s f i KE PE KE PE + = + , or 2 1 0 0 60 J 2 mv + = + , so  ( 29 -3 2 60 J 50 10  kg v = = × 49  m s 13.8 (a) 230 N 0.400 m max max F k x = = = 575  N m (b) ( 29 ( 29 2 2 1 1   575  N m 0.400 2 2 s work done PE kx = = = = 46.0 J
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
13.16 (a) 0 at  KE x A = = , so  2 1 0 2 s E KE PE kA = + = + , or the total energy is ( 29 ( 29 2 2 1 1 250  N m 0.035 m 2 2 E kA = = = 0.15 J (b) The maximum speed occurs at the equilibrium position where  0 s PE = . Thus,  2 1 2 max E mv = or ( 29 2 250  N m 0.035 m 0.50 kg max E k v A m m = = = = 0.78  m s (c) The acceleration is  F kx a m m Σ - = = . Thus,  max a a =  at  max x x A = - = - . ( 29 ( 29 250  N m 0.035 m 0.50 kg max k A k a A m m - - = = = = 2 18 m s 13.26 (a) At  0 t = ( 29 ( 29 0.30 m cos 0 x = = 0.30 m , and at  0.60 s t = , ( 29 ( 29 ( 29 ( 29 0.30 m cos  rad s 0.60 s 0.30 m cos 0.20  rad 3 x π π = = = 0.24 m (b) ( 29 ( 29 0.30 m 1 max A x = = = 0.30 m (c) ( 29 0.30 m cos 3 x t π =  is of the form  ( 29 cos x A t ϖ =  with an angular frequency of   rad s 3 π ϖ = . Thus,  3 2 2 f π ϖ π π = = = 1  Hz 6 (d) The period is  1 T f = = 6.0 s
Image of page 2
13.30 (a) The height of the tower is almost the same as the length of the pendulum. From  2 T L g π = , we obtain ( 29 ( 29 2 2 2 2 2 9.80  m s 15.5 s 4 4 gT L π π = = = 59.6 m (b) On the Moon, where  2 1.67  m s g = , the period will be 2 59.6 m 2 2 1.67  m s L T g π π = = = 37.5 s 13.34 The apparent acceleration of gravity is the vector sum of the actual  acceleration of gravity and the negative of the elevator’s acceleration. To see 
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
Image of page 4
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