242_Physics ProblemsTechnical Physics

# 242_Physics ProblemsTechnical Physics - 244 P8.66 Potential...

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244 Potential Energy P8.66 Case I: Surface is frictionless 1 2 1 2 22 mv kx = k mv x == = × 2 2 2 2 2 500 120 10 720 10 .. . kg m s m ±Nm 2 bg b g Case II: Surface is rough, µ k = 0300 . 1 2 1 2 mv kx mgx k =− 1 2 7 20 10 10 0 300 5 00 9 80 10 0923 1 2 1 . . . . kg 2 Nm m kg ms m 2 v v = −− ej e j a f e j *P8.67 (a) KU g A g B += + 0 1 2 0 2 + mgy mv AB vg y BA = 2 298 63 111 ... m 2 (b) a v r c = 2 2 19 6 . . . m ms up 2 (c) Fm a yy =+ = nm gm a Bc n B = × 76 9 8 19 6 2 23 10 3 kg m s m s N up . (d) WF r × ° = × cos . . cos . θ 223 10 0450 0 101 10 33 N m J af (e) W KU g B g D ++ = + 1 2 0 1 01 10 1 2 1 2 76 11 1 1 01 10 1 2 76 76 9 8 6 3 5 70 10 4 69 10 2 76 514 23 2 2 32 mv mv mg y y v v BD D B D D × = + = + ×− × . . . . J kg m s J kg kg m s m J J kg 2 (f) g D g E + where E is the apex of his motion 1 2 00 2 mv mg y y DE D +=+ v g ED D −= = = 2 2 2 135 . . . m 2 (g) Consider the motion with constant acceleration between takeoff and touchdown. The time
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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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