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A plank of mass m1=8 kg with a bar of mass m2=2 kg placed on its rough surface, lie on a smooth floor of elevator ascending with an acceleration g/4. The coefficient of friction is mu=1/5 between m1 and m2. A horizontal force F=30N is applied to the plank. Then the acceleration of a bar and the plank in the reference frame of elevator are

A plank of mass m1=8 kg with a bar of mass m2=2 kg placed on its rough surface, lie on a smooth floor of elevator ascending with an acceleration g/4. The coefficient of friction is mu=1/5 between m1 and m2. A horizontal force F=30N is applied to the plank. Then the acceleration of a bar and the plank in the reference frame of elevator are

Grade:11

2 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
let the frictional force between plank and bar is f
f=µN

N=pseudo force+weight of the bar
=20+2g/4=25N

f=25x1/5 = 5N
acceleration of the bar = force on bar/mass of bar = frictional force/mass
=5/2
=2.5m/s2

acceleration of the plank=30-5/8
=25/8 m/s2
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

 
Poojitha
16 Points
3 years ago
  Here as given in the question,
  m1=8kg,m2=2kg ,
  force F acting on the block=30N
  and coefficient of friction between the blocks
  is 1/5;
  Then ,let "a" be the acceleration with which the 
  elevator is moving = g/4,and a2 is acc. of upper block,a1 is acc. of lower block.
  Then;let f be the friction between the two blocks,
  then,for upper block,
   f=coefficient of friction ×(m2)(g+a) = m2×a (let it be 1)
   For lower block ,
   Force F and friction acts in opposite directions
   So,F-f= m2× a1(let it be 2)
   by solving 1,as a=g/4,so 
   We get a2 = 2.5, and a1 =25/8
   
 

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