228_Physics ProblemsTechnical Physics

228_Physics ProblemsTechnical Physics - 230 Potential...

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

View Full Document Right Arrow Icon
230 Potential Energy P8.33 (a) Km v v m v fi i =− = = 1 2 1 2 160 22 2 ej J (b) Um g = 300 735 .s i n . . m J a f (c) The mechanical energy converted due to friction is 86.5 J f == 86 5 28 8 . . J 3.00 m N (d) fnm g kk ° = µµ cos . . 30 0 28 8 N µ k = ° = 28 8 980 0679 . .c o s . . N 5.00 kg m s 2 bg FIG. P8.33 P8.34 Consider the whole motion: KU E K U ii f f ++ = + mech (a) 0 1 2 0 80 0 9 80 1 000 50 0 800 3 600 200 1 2 80 0 784 000 40 000 720 000 1 2 80 0 224000 80 0 24 5 11 22 2 2 2 +− = + −− = = mgy f x f x mv v v v if f f f ∆∆ .. . . . . . kg m s m N m N m kg J J J k g J kg ms 2 a fa f a f (b) Yes this is too fast for safety. (c) Now in the same energy equation as in part (a), x 2 is unknown, and xx 12 1000 m: 784 000 50 0 1 000 3 600 1 2 80 0 5 00 784 000 50 000 3 550 1 000 733 000
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online