CHAPTER 4 - 5

# CHAPTER 4 - 5 - (43kg(9.80m/s)F 2 legs...

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F N y F G1 F G2 F T F T (43 kg)(9.80 m/s 2 ) –  F legs  = (43 kg)(– 49 m/s 2 ), which gives        F legs  = 2.5 × 10 3  N up . 25. ( a ) If we assume that he accelerates for a time  t 1  over the first 45 m and reaches a top speed  of  v , we have x 1  =  ! ( v 0  +  v ) t 1  =  ! vt 1   ,   or    t 1  = 2 x 1 / v  = 2(45 m)/ v  = (90 m)/ v . Because he maintains this top speed for the last 55 m, we have t 2  = (55 m)/ v . Thus the total time is  T  =  t 1  +  t 2  =  (90 m)/ v + (55 m)/ v  = 10.0 s. When we solve for  v , we get  v  = 14.5 m/s; so the acceleration time is  t 1  = (90 m)/(14.5 m/s) = 6.21 s. We find the constant acceleration for the first 45 m from a  = ? v /? t  = (14.5 m/s – 0)/(6.21 s) = 2.34 m/s 2 . We find the horizontal force component that will produce this acceleration from F  =  ma  = (62 kg)(2.34 m/s

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## This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at Berkeley.

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CHAPTER 4 - 5 - (43kg(9.80m/s)F 2 legs...

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