Example 5.9 What is the acceleration of the block and trolley system shown in a Fig. 5.12(a), if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? (Take g = 10 m s-2). Neglect the mass of the string.

Kinetic coefficient of  friction μk=0.04 given. From the diagram the equations of motion of the system network is given by,30-T= M1a            or  30-T=3×a  -----1  andT-fk=M2a------2.In the above equations M1 and M2 are the masses of block and trolley.fk=μkR       and  R=M2×g=20×g=20×10=200N.Therefore fk=0.04×200=8N. So equation 2 becomes,T-8=20×a----3.Adding equations 1 and 3,30-8=23×a.      or     a=2223.      or     a=0.956 ms-2.And  T= 20×a+8.        Or       T=20×2223+8=440+18423=62423.  Or        T=27.1 NTherefore acceleration a=0.956 ms-2 and  tension T= 27.1 N.

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