40. (a) The initial direction of motion is taken to be the +xdirection (this is also the direction of GE). We use vvaxfi222−=∆with vf= 0 and GGGaFm eEme==−to solve for distance ∆x: ∆xvamveEiei=−=−−=−×××=×−−−31192911 102 160 10712 10...kg 5.00 10 m sC 1.00 10 N Cm.623chchch(b) Eq. 2-17 leads to txvxvi=××−−∆∆avgmmss.22 712 10500 10285 10268...ch(c) Using ∆v2= 2a∆xwith the new value of ∆x, we find
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