Principle of Work and Energy for Particles in a Plane  (example continued)

 

 

Strategy:  Apply the principle of work and energy to determine the speed of the blocks

after B has dropped 2 ft.   Define the system to be block A + the pulley + block B.   The

work done on the system  W1-2,  from position 1 to position 2 equals the change in kinetic

energy of the system  T2 – T1  such as:

                                                                    W1-2  =  T2  -  T1

                              

                                                                         

 

 

Show the forces that do work during displacement, d, of the system from position (1) to

position (2).  See the figure below.   

                                       

                                                  

 

 

The work done on the system,  W1-2  =  WB d – μ WA d

 

 

The change in kinetic energy    T1-2  =    T2 – T1 =  ½ ma v22  +  ½ mbv22  -  0

 

                 WB d – μ WA d   =   ½ ma v22 +  ½ mb v22 

 

 

       v22  =      2 ( WB  – μ WA ) d  / (ma + mb )  =  2g (WB  – μ WA )(2) / (WA + WB)

 

     v22  =  √ [2g (WB  – μ WA )(2) / (WA + WB)]  =  √ [ 64.4 (16.1 – 6.44)(2)/(48.3) ]

 

     v2  =  5.075 ft/sec       (result)     v1  =  v2  since cord is inextensional.

 

 

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