2014_SP08_Sol09

# 2014_SP08_Sol09 - Work 11-1 11.6 Solve(a(b W F ^ ^ ^ ^ W F...

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Work 11-1 11.6. Solve: (a) x Éº ª* ( 5.0 4.0 ) N (3.0 ) m ( 15.0 12.0 ) J= 15.0 J. W F r i j i i i j i = ⋅∆ = - + = - ⋅ + - r r r (b) x É ≡♠ ( 5.0 4.0 ) N ( 3.0 ) m (15.0 12.0 ) J= 12.0 J. W F r i j j i j j j = ⋅∆ = - + ⋅ - = ⋅ - - r r r 11.13. Model: Use the work-kinetic energy theorem to find velocities. Visualize: Please refer to Figure Ex11.13. Solve: The work-kinetic energy theorem is f i f x 2 2 f i i f x 2 2 2 f f 0 m 1 1 area under the force curve from to 2 2 1 1 1 (0.500 kg)(2.0m/s) 1.0 J area from 0 to 2 2 2 x x x K mv mv W F dx x x mv mv F dx x = - = = = - = - = = 2 f f 2 f f 2 f f 1 At =1 m: (0.500 kg) 1.0 J 12.5 J 7.35 m/s 2 1 At 2 m: (0.500 kg) 1.0 J 20 J 9.17 m/s 2 1 At 3 m: (0.500 kg) 1.0 J 22.5 J 9.70 m/s 2 x v v x v v x v v - = = = - = = = - = = 11.40. Model: Model the rock as a particle, and apply the work-kinetic energy theorem. Visualize: Solve: (a) The work done by Bob on the rock is 2 2 2 2 Bob 1 0 1 1 1 1 1 (0.50 kg)(30 m/s) 225 J 2 2 2 2 W K mv mv mv = ∆ = - = = = (b) For a constant force, Bob Bob Bob Bob / 225 N W F x F W x = ∆ ⇒ = ∆ = . (c) Bob’s power output is Bob Bob rock P F v = and will be a maximum when the rock has maximum speed. This is just as he releases the rock with v rock = v 1 = 30 m/s. Thus, max Bob 1 6750 W 6.75 kW. P F v = = = 11.48. Model: Model the two blocks as particles. The two blocks make our system.

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Work 11-2 Visualize: We place the origin of our coordinate system at the location of the 3.0 kg block. Solve:
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2014_SP08_Sol09 - Work 11-1 11.6 Solve(a(b W F ^ ^ ^ ^ W F...

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