197_Physics ProblemsTechnical Physics

# 197_Physics - Chapter 7 P7.26 e 199 j v i = 6.00 i 2.00 j = m s(a 2 2 vi = vix viy = 40.0 m s Ki =(b b ge 1 1 mvi2 = 3.00 kg 40.0 m 2 s 2 = 60.0 J

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Chapter 7 199 P7.26 vi j i =−= 600 200 . ± . ± ej ms (a) vv v ii x i y =+ = 22 40 0 . ms Km v == = 1 2 1 2 300 400 2 .. . kg m s J bg (b) j f 800 . ± . ± v ff f 2 64 0 16 0 80 0 = =+= ... m s KK K m v v fi =− = − = = 1 2 2 80 0 60 0 60 0 a f . . J P7.27 Consider the work done on the pile driver from the time it starts from rest until it comes to rest at the end of the fall. Let d = 5.00 m represent the distance over which the driver falls freely, and h = 012 . m the distance it moves the piling. WK =∆ : W W mv mv gravity beam +=− 1 2 1 2 so mg h d F d af di + ° = cos cos 01 8 0 0 0 . Thus, F mg h d d = + × 2100 980 512 0120 878 10 5 kg m s m m N 2 . . . The force on the pile driver is upward . P7.28 (a) m v W f = = = 1 2 0 2 (area under curve from x = 0 to x = 500 . m ) v m f = 7 5 0 194 area J kg a f . . . (b) m v W f = = = 1 2 0 2 (area under curve from x = 0 to x = 10 0 . m ) v m f = 2 2 5 335 area J kg a f . . . (c) m v W f = = = 1 2 0 2 (area under curve from
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## 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 .

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