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Unformatted text preview: Chapter 7 HW Solutions 16. The forces are all constant, so the total work done by them is given by W F x = net ∆ , where F net is the magnitude of the net force and ∆ x is the magnitude of the displacement. We add the three vectors, finding the x and y components of the net force: net 1 2 3 net 2 3 sin50.0 cos35.0 3.00 N (4.00 N)sin35.0 (10.0 N)cos35.0 2.13N cos50.0 sin35.0 (4.00 N) cos50.0 (10.0 N)sin35.0 3.17 N. x y F F F F F F F =  ° + ° =  ° + ° = =  ° + ° =  ° + ° = The magnitude of the net force is 2 2 2 2 net net net (2.13 N) (3.17 N) 3.82 N. x y F F F = + = + = The work done by the net force is net (3.82 N)(4.00m) 15.3 J W F d = = = where we have used the fact that d F net  (which follows from the fact that the canister started from rest and moved horizontally under the action of horizontal forces — the resultant effect of which is expressed by F net ). 32. Hooke’s law and the work done by a spring is discussed in the chapter. We apply workkinetic energy theorem, in the form of ∆ K W W a s = + , to the points in Figure 738 at x = 1.0 m and x = 2.0 m, respectively. The “applied” work W a is that due to the constant force P . 2 2 1 4 J (1.0 m) (1.0 m) 2 1 (2.0 m) (2.0 m) 2 P k P k = = (a) Simultaneous solution leads to P = 8.0 N....
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 Spring '10
 MichaelGorman
 Physics, Energy, Force, Work, WorkKinetic Energy Theorem, Fnet

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