This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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. Hookes 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....
View Full
Document
 Spring '10
 MichaelGorman
 Physics, Force, Work

Click to edit the document details