two blocks of  diff masses r placed over each other on a rough horizontal table.the mass of d lower block is 10 kg nd dat of d upper mass is 5kg.the friction coefficient b/w d 2 blocks is 0.1 nd dat b/w d table nd d lower block is also 0.1.a force F is applied on mass 10 kg towards right.

find d acceleration of d 2 blocks when 1)F=100N, 2).F=30N and3)F=20N

WHEN force =100n

  now here frictional force on block  = (10+5)*10*0.1= 15n

so the block can move hence by FBD of 10kg block

  EQUATION=  F-15-5=100- 20 = 80

    accn = 80/10=8

for smaller block accn by pseudo force  = 5*8-0.1*50 = 40-5 =35/5=7m/s^2

so ans is 8m/s^2 and 7m/s^2

is it right ?????   so that i can answer other parts

is abhi right? . plz slv d othr 2, nd gv me d ans. i hv slvd dem but just wnt 2 cnfrm b'cauz i d'nt hv d ans.

NOW  WHEN FORCE = 30N

               FBD FOR BIGGER BLOCK

              = 30-20=10N

so accn = 10/10=1m/s^2

 so pseudo force  on m  = 5N 

BUT WE CAN SEE THAT = 5N  SAME  so the smaller block will move with M block

  so they both move together   accn = 30-15=15/15 = 1m/s^2  for both

 and for  F = 20N  the blocks will not move so accn =0m/s^2

PLZ RATE YAAR!!!!!!!!

Varun you are totally wrong in 1st part 

where is the pseudo force man!!!!!!!!!!

you have not taken it into consideration

second case for the lower block 30-f1-f2=10*a =>30-15-5=m10*a => a=1 (lower)

now for the upper... 5=5*a => a=1 so both move together

now in the third q ABHI i would like u to post the sol for the third part in ur method...bcos i still am not very clear how to workout this case from relative frame however i will post the sol according to my method..... do the same F-20 =10*a u get a=0 now this is relative acc.... relative acc is 0 means they dont slip but they move together

so 20 -15 =15*a => a= 5/15 =1/3........ABHI i am waiting for ur method....

WELL i think you have only mentioned the " respect to ground method " 

firstly you have proved that there relative vel is 0

then      by   you have got the ans 1/3m/s^2

which is respect to ground only            

SO YOUR METHOD IS WITH RESPECTIVE TO GROUND ONLY

 but he has stated that accn  lower block is 8m/s^2

and of upperblock is 1m/s^2!!!!!!!!!!!

ya thats what i have done but only by using ground that is my method i prove that relative vel is 0. and use that but what i wanted to know is that how will u do the q without proving rel. vel is 0 or by proving it only relative frame......

I HAVE THAT METHOD ONLY  !!!!!!!!