chapter 6 problem 07 Chapter 06, Problem 07 Consult Interactive LearningWare 6.1

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chapter 6 problem 07

Chapter 06, Problem 07 Consult Interactive LearningWare 6.1 for background pertinent to this these motions, the lift force L acts perpendicular to the displacement s, which has the same magnitude, 3.56 x 103 m, in each case. The engines of the plane exert a thrust T, which points in the direction of the displacement and has the same magnitude during the dive and the a m The weigh l om the na e san agnitude of 20·0·i n on net work is performed due to the combined action of the forces L, T, and W. Find the difference between the net work done during the dive and the climb. s problem. The drawing shows a plane diving toward the ground and then climbing back upward. During each of 115 75 (b) Climb a) Dive W Number Units the tolerance is +/-2% By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor. Question Attempts: 0 of 5 used SAVE FOR LATER 23 PM to search

Explanation / Answer

07. form the given diagram and the data
   thrust = T
   lift = L
   weight = W

   so while diving
   net work done for a displacement s
   W = Ts + Wcos(75)*s

   while climbing
   net work done
   W = Ts + wcos(115)s

   now, W = 6.9*10^4 N
   and s = 3.56*10^3 m

   so difference in work done during dive and climb = Ts + Wcos(75)s - Ts - Wcos(115)s
   dW = 6.9*10^4*3.56*10^3(cos(75) - cos(115)) = 16.73*(10^7) J

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