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In figure block 1 is one fourth of length l of block 2of mass also one fourth. No friction exist betweenblock 2 and surface on which it rests. Coefficient offriction is µk between 1 and 2. Find the distanceblock 2 moves when only half of block 1 is still onblock 2. Block 1 and block 3 have same mass.

In figure block 1 is one fourth of length l of block 2of mass also one fourth. No friction exist betweenblock 2 and surface on which it rests. Coefficient offriction is µk between 1 and 2. Find the distanceblock 2 moves when only half of block 1 is still onblock 2. Block 1 and block 3 have same mass.

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Grade:11

1 Answers

faizan
13 Points
7 years ago
For the block 3 ,
let a be its accelaration . so net forces acting on it are 1) its weight downward  2) tension T in the string upward ,
equating this we have
  mg-T = ma -------(1)
for bock 1
net forces are :-
1) tension T forward   2 )friction fs
since 1 and 3 are connected by a string thy share same accelaration .   equating this
T-f=ma  , but fs=ukmg so T-ukmg = ma -------(2)   
adding eqn 1 and 2
we have mg(1-uk)=2ma   or g/2(1-uk)=a   
now finally for bloack 3 net force is fs in forward direction so if its accelaration is f/m or mguk/M ----3 
but in time t distace travelled by bloack 1 is 7/8L (L isthe length of bnlock 2)   so using s = 1/2at2 
time taken by block 2 to travel 7/8L is t2 = 7Lmuk /4(1-u) -----4 
finally distance moved by block 2 in time t is 7Lmuk/4M(1-uk) … the answer is complete

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