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Chapter 11

# Chapter 11 - 11.1 Visualize Please refer to Figure EX11.1...

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11.1. Visualize: Please refer to Figure EX11.1. Solve: (a) cos (4)(5)cos40 15.3. A B AB α = = ° = G G (b) cos (2)(4)cos120 4.0. C D CD α = = ° = − G G (c) cos (3)(4)cos90 0. E F EF α = = ° = G G

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11.2. Visualize: Please refer to Figure EX11.2. Solve: (a) cos (3)(4)cos110 4.1. A B AB α = = ° = − G G (b) cos (4)(5)cos180 20. C D CD α = = ° = − G G (c) cos (4)(3)cos30 10.4. E F EF α = = ° = G G
11.3. Solve: (a) (3)( 2) ( 4)(6) 30. x x y y A B A B A B = + = + − = − G G (b) (2)(6) (3)( 4) 0. x x y y A B A B A B = + = + = G G

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11.4. Solve: (a) (4)( 3) ( 2)( 2) 16. x x y y A B A B A B = + = + − = − G G (b) ( 4)( 1) (2)( 2) 0. x x y y A B A B A B = + = − + = G G
11.5. Solve: (a) The length of A G is ( ) ( ) 2 2 3 4 25 5. A A A A = = = + − = = G G G The length of B G is ( ) ( ) 2 2 2 6 B = + = 40 2 10. = = Using the answer 30 A B = − G G from Ex11.3(a), ( ) ( ) ( ) 1 cos 30 5 2 10 cos cos 0.9487 162 A B AB α α α = ⇒ − = = = ° G G (b) From EX11.3 (b), 0. A B = G G Thus cos 0 90 α α = = ° Assess: In part (a) the vectors are nearly pointing in opposite directions, while in part (b) the vectors are perpendicular.

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11.6. Solve: (a) The length of A G is ( ) ( ) 2 2 4 2 20. A A A A = = = + = G G G The length of B G is ( ) ( ) 2 2 3 2 B = + − = 13. Using the answer 16 A B = − G G from EX11.4(a), ( ) ( ) ( ) 1 cos 16 20 13 cos cos 0.9923 173 A B AB α α α = ⇒ − = = = ° G G (b) From EX11.4(b), 0. A B = G G Thus cos 0 90 α α = = ° Assess: In part (a) the vectors are nearly pointing in opposite directions, while in part (b) the vectors are perpendicular.
11.7. Solve: (a) ˆ ˆ ˆ ˆ ˆ ˆ ˆ (6.0 3.0 ) (2.0 ) N m (12.0 3.0 ) J 12.0 J. W F r i j i i i j i = ⋅Δ = = = G G (b) ˆ ˆ ˆ ˆ ˆ ˆ ˆ (6.0 3.0 ) (2.0 ) N m (12.0 6.0 ) J 6.0 J. W F r i j j i j j j = ⋅Δ = = = − G G

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11.8. Solve: (a) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ( 4.0 6.0 ) N (3.0 ) m ( 12.0 12.0 ) J= 12.0 J. W F r i j i i i j i = ⋅Δ = − = − + G G G (b) ˆ ˆ ˆ ˆ ( 4.0 6.0 ) N ( 3.0 2.0 ) m (12.0 12.0) J=0 J. W F r i j i j = ⋅Δ = − ⋅ − + = G G G
11.9. Model: Use the work-kinetic energy theorem to find the net work done on the particle. Visualize: Solve: From the work-kinetic energy theorem, ( ) 2 2 2 2 2 2 1 0 1 0 1 1 1 1 (0.020 kg)[(30 m/s) ( 30 m/s) ] 0 J 2 2 2 2 W K mv mv m v v = Δ = = = − − = Assess : Negative work is done in slowing down the particle to rest, and an equal amount of positive work is done in bringing the particle to the original speed but in the opposite direction.

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11.10. Model: Work done by a force F G on a particle is defined as , W F r = ⋅Δ G G G where r Δ G is the particle’s displacement. Visualize: Solve: (a) The work done by gravity is 2 g G ˆ ˆ ( ) N (2.25 0.75) m (2.0 kg)(9.8 m/s )(1.50 m) J 29 J W F r mgj j = ⋅Δ = − = − = − G G (b) The work done by hand is H hand on book . W F r = ⋅Δ G G As long as the book does not accelerate, hand on book earth on book 2 H ˆ ˆ ( ) ˆ ˆ ( ) (2.25 0.75) m (2.0 kg)(9.8 m/s )(1.50 m)=29 J F F mgj mgj W mgj j = − = − − = = = G G
11.11. Model: Model the piano as a particle and use , W F r = ⋅Δ G G where W is the work done by the force F G through the displacement . r Δ G Visualize: Solve: For the force G : F G 4 G ( )( )cos0 (2500 N)(5.00 m)(1) 1.250 10 J g W F r F r F r = ⋅Δ = ⋅Δ = Δ ° = = × G G G G For the tension 1 : T G 3 1 1 ( )( )cos(150 ) (1830 N)(5.00 m)( 0.8660) 7.92 10 J W T r T r = ⋅Δ = Δ ° = = − × G G For the tension 2 : T G 3 2 2 ( )( )cos(135 ) (1295 N)(5.00 m)( 0.7071) 4.58 10 J W T r T r = ⋅Δ =

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Chapter 11 - 11.1 Visualize Please refer to Figure EX11.1...

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