# Chapter11 - 11.1 Visualize:r Please refer to Figure Ex11.1...

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11.1. Visualize: Please refer to Figure Ex11.1. Solve: (a) r r A B AB = = ° = cos ( )( )cos . . α 4 5 40 15 3 (b) r r C D CD = = ° = − cos ( )( )cos . . α 2 4 120 4 0 (c) r r E F EF = = ° = cos ( )( )cos . α 3 4 90 0

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11.2. Visualize: Please refer to Figure Ex11.2. Solve: (a) r r A B AB = = ° = − cos ( )( )cos . . α 3 4 110 4 1 (b) r r C D CD = = ° = − cos ( )( )cos . α 4 5 180 20 (c) r r E F EF = = ° = cos ( )( )cos . . α 4 3 30 10 4
11.3. Solve: (a) r r A B A B A B x x y y = + = + − = − ( ( ( )( ) . 3) ) 2 4 6 30 (b) r r A B A B A B x x y y = + = + = ( ( ( )( ) 2) ) 6 3 4 0 .

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11.4. Solve: (a) r r A B A B A B x x y y = + + − = = 5) ) ( ( ( )( ) . 2 3 2 16 (b) r r A B A B A B x x y y = + = + − = − ( ( ( )( ) . 4) ) 3 2 6 24
11.5. Solve: (a) W F r i j i i i j i = = = = r r ( . ˆ . ˆ ) ( . ˆ ) ( . ˆ ˆ . ˆ ˆ ) . 6 0 3 0 2 0 12 0 3 0 12 0 N m J J. (b) W F r i j j i j j j = = = = − r r ( . ˆ . ˆ ) ( . ˆ ) ( . ˆ ˆ . ˆ ˆ ) . 6 0 3 0 2 0 12 0 6 0 6 0 N m J J.

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11.6. Solve: (a) r r r W F r i j i i i j i = = − + = − + ( . ˆ . ˆ ) ( . ˆ ) ( . ˆ ˆ . ˆ ˆ ) 5 0 4 0 3 0 15 0 12 0 N m J = 15.0 J. (b) r r r W F r i j j i j j j = = − + ⋅ − = ( . ˆ . ˆ ) ( . ˆ ) ( . ˆ ˆ . ˆ ˆ ) 5 0 4 0 3 0 15 0 12 0 N m J = 12.0 J.
11.7. Model: Use the work-kinetic energy theorem to find the work done on the particle. Visualize: Solve: From the work-kinetic energy theorem, W K mv mv m v v = = = ( ) = − − = 1 2 1 2 1 2 1 2 0 1 2 0 2 1 2 0 2 2 2 2 ( )[( ) ( ) ] 0.020 kg 30 m/s 30 m/s J 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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11.8. Model: Work done by a force r F on a particle is defined as r r r W F r = ⋅ ∆ , where r r is the particle’s displacement. Visualize: Solve: (a) The work done by gravity is W w r mgj j g 2 N m 2.0 kg 9.8 m/s 1.50 m J 29.4 J = = − = − = − r r ( ˆ ) ( . . ) ˆ ( )( )( ) 2 25 0 75 (b) The work done by hand is W F r H hand on book = r r . As long as the book does not accelerate, r r F F mgj mgj W mgj j hand on book earth on book H 2 m 2.0 kg 9.8 m/s 1.50 m = 29.4 J = − = − − = = = ( ˆ ) ˆ ( ˆ ) ( . . ) ˆ ( )( )( ) 2 25 0 75
11.9. Model: Model the piano as a particle and use W F r = r r , where W is the work done by the force r F through the displacement r r . Visualize: Solve: For the force r w : W F r w r w r = = = ° = = r r r r ( )( )cos ( )( . )( ) , 0 2500 5 0 1 12 500 N m J For the tension r T 1 : W T r T r = = ° = = − r r 1 1 150 1830 5 0 0 8660 7920 ( )( )cos( ) ( )( . )( . ) N N J For the tension r T 2 : W T r T r = = ° = = − r r 2 2 135 1295 4580 ( )( )cos( ) ( )( )( ) N 5.0 m 0.7071 J Assess: Note that the displacement r r in all the above cases is directed downwards along ˆ . j

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11.10. Model: Model the crate as a particle and use W F r = r r , where W is the work done by a force r F on a particle and r r is the particle’s displacement. Visualize: Solve: For the force r f k : W f r f r = = ° = = − r r k k 500 N) 3.0 m) ) J ( )cos( ) ( ( ( 180 1 1500 For the tension r T 1 : W T r T r = = ° r r 1 1 20 0 9397 ( )( )cos ( ( ( . = 326 N) 3.0 m) ) = 919 J For the tension r T 2 : W T r T r = = ° = = r r 2 2 30 223 579 ( )( )cos ( )( ) ) N 3.0 m (0.866 J Assess: Negative work done by the force of kinetic friction ( ) r f k means that 1500 J of energy has been transferred out of the crate.
11.11.

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