Physics, asked by Utsavsterbon, 1 year ago

Two constant horizontal forces F1 and F2 are acting on blocks A and B. At an instant acceleration of block B is 4 m/s2 in direction of F1 and acceleration of block A with respect to block B at this moment is :

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Answered by kvnmurty
5
Initially the spring extends in length, then its restoration force on either end will be same and equal to  - k x, where k = force constant of the spring and x is the extension of the spring.  Hence block A is pulled towards right with the same spring force.

Equations of dynamics of the blocks are 
 
      F1 - k x  =  M_B * a_B
      k x - F2  =  M_A * a_A

    F1 - F2 = M_B * a_B + M_A * a_A 
     100 - 70  = 5 * 4 + 2 * a_A
    
          a_A = 5 m/sec²



kvnmurty: please click on thank you link and select best answer
Utsavsterbon: sir can you please explain where did the first equation come from?
Answered by Yeshwanth1245
0

Initially the spring extends in length, then its restoration force on either end will be same and equal to  - k x, where k = force constant of the spring and x is the extension of the spring.  Hence block A is pulled towards right with the same spring force.

Equations of dynamics of the blocks are  

 

     F1 - k x  =  M_B * a_B

     k x - F2  =  M_A * a_A

   F1 - F2 = M_B * a_B + M_A * a_A  

    100 - 70  = 5 * 4 + 2 * a_A

   

         a_A = 5 m/sec²

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