Example 5 9 What is the acceleration of the block and trolley system shown in a Fig 5 12(a), - Physics - Laws Of Motion

Question

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

Ved Prakash Lakhera · Accepted Answer

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

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.