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# Two blocks A and B of mass 6kg and3g respectively are placed on asmooth surface of anelevator in such a way that B isplaced above A.The coefficient offriction between the twoblocks is 1/4 and the elevator isascending up with accelerationg/3.A horizontal force 10Nis applied on A. Find theaccelerations of the blocks w.r.t. theelevator.

71 Points
7 years ago
First draw the diagram by yourself according to the the given conditions and draw FBD. Mass of A= 6Kg, Mass of B= 3Kg As elevator is accelerating upwards with acceleration of g/3,value of Normal Reaction(N) will change N-Mg=Mg/3 N=4Mg/3 Hence value of N between 2 blocks will be = 4*3*9.8/3= 39.2Newton Now common acceleration of 2 blocks in horizontal direction = 10/(3+6) =10/9 m/s2 *Now check whether there will be slipping between 2 blocks or not Force required by B to move with common acceleration in horizontal direction = 3*10/9 = 10/3 Newton Maximum friction between blocks = µ*N = 1/4*39.2 = 9.8 Newton As 9.8>10/3, There will be no slipping between the blocks. They will move with common acceleration in horizontal direction that is 10/9 m/s2 Hence acceleration of both blocks w.r.t. the elevator is = (10/9 i cap + g/3 j cap) m/s2....
Harsh Srivastava
34 Points
7 years ago
Sorry my final answer is w.r.t Earth. Hence acceleration of both blocks w.r.t. the elevator is = (10/9 i cap ) m/s2....
18 Points
7 years ago
Is it 3g or 3kg.....
Nirmal Singh.
7 years ago

Thanks & Regards,
Nirmal Singh