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CHAPTER 4 - 11

# CHAPTER 4 - 11 - ThisisalsoF(Newtonsthirdlaw 32(e ,weget...

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This is also  F 32  (Newton’s third law).  ( e ) When we use the given values, we get a  =  F /( m 1  +  m 2  +  m 3 ) = (96.0 N)/(12.0 kg + 12.0 kg + 12.0 kg) =         2.67 m/s 2 . F net1  =  m 1 a  = (12.0 kg)(2.67 m/s 2 ) = 32.0 N. F net2  =  m 2 a  = (12.0 kg)(2.67 m/s 2 ) = 32.0 N. F net3  =  m 3 a  = (12.0 kg)(2.67 m/s 2 ) = 32.0 N. Because the blocks have the same mass and the same acceleration, we expect F net1  =  F net2  =  F net3  = 32.0 N . For the forces between the blocks we have F 21  =  F 12  =  F   –  m 1 a  = 96.0 N – (12.0 kg)(2.67 m/s 2 ) =        64.0 N . F 32  =  F 23  =  F   –  m 1 a  –  m 2 a  = 96.0 N – (12.0 kg)(2.67 m/s 2 ) – (12.0 kg)(2.67 m/s 2 ) =        32.0 N . 47.                  ( a ) ( b ) If the system is released from rest, the blocks will  have the same acceleration in the directions  indicated on the diagram.  We write ? F  =  m from

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