sol to assignment5

# sol to assignment5 - Auswges ‘67 Crt 5...

• Homework Help
• PresidentHackerCaribou10582
• 1

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Auswges ‘67? Crt. 5 ﬂ-SSIeM,ME7-JT‘ 5.2 w= mg = (20.0 kg)(9.80 nth/32) and 196 N = the applied force to lift the bucket at constant w= (PcosB)s= 600x103 J Fs= (196 N) s yielding, s= 30 6 In. , 5.1 2 At the top of the arc, V; = 0, and vx = vex = 40 (203300“ = 34.64 m/s. Therefore v2 = V2}; + v y = (34.64 iii/3F, and 1 KE= 5 va =§ (0.150)(34.64)2 = 90.0 J. 5.2 1 (a) We take the zero level of potential energy at the lowest point of the arc. When the suing is held horizontal initially, the initial position is 2 m above the zero level. Thus, PE= mgy= wy= (40 N)(2 m) =- 80 J. (b)Promthesketch, weseethatat ----"-----“-0 an angle of 30° the ball is : (2.0 m)(1 - c0330“) above the lowest point of the arc. Thus, P134 (40 N)(2.0 m)(1 - 00330“) = 11 J. (c) The zero level has been selected at the lowest point the arc. Therefore,. PE = O at this point. 1 + ”1353* 2 mvfz + mgyr to obtain 0+(0.400kg)g(5.00m)= 2(0.400kg)v132+0. Fromthis, VB=990m/S. (b) At int f0 C ,with theistarting point at A, we again use mez+mgﬁ= 2me2+:ﬂgy1= andobtain 5.28 (a) Liming m the the mass of the projectile, k be the spring constant, C! be 0.120 in, and H== 20.0 m, we have, using conservation of energy 2_mg_ H_ 2.!0 0211980190. 0') — 2,. _ a chf mgH, or k— _2g_ (0120): 544N/n1. )Here we have ink? =lrngd+-'21,mv2 so that 2 vat/I‘d ~ 2 gd— SMSOlezol—JLWJ 2(9.80)(O.120) =19.7 m/s. At 5.43 Pag’ﬂ'gALtsQQQ-JA—mmg = (1.2): 106)(9.80)(50) ' =5.88x108W=590 MW. ...
View Full Document

• Fall '01
• Shapiro

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern