HW3 Solutions

# A with k mv i 1 2 2 16 j we have 3 1 1 2 2 3 40 j x x

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(a) With K mv i 1 2 0 2 16 J, we have 3 0 0 1 1 2 2 3 4.0 J x x x K K W W W        so that K 3 (the kinetic energy when x = 3.0 m) is found to equal 12 J. (b) With SI units understood, we write 3 as ( 4.0 N)( 3.0 m) f x x x f W F x x      and apply the work-kinetic energy theorem: K K W K x x x x x f f f f     3 3 12 4 30 ( )( . ) so that the requirement 8.0 J xf K leads to x f 4 0 . m. (c) As long as the work is positive, the kinetic energy grows. The graph shows this situation to hold until x = 1.0 m. At that location, the kinetic energy is 1 0 0 1 16 J 2.0 J 18 J. x K K W  
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