Question

`a block of mass 2 kg placed on another block of 3 kg kept on a smooth horizontal table coefficient of friction between two block is 1. a horizontal force of 25N is applied horizontally 2kg block then force between two blocks is

Answer 1

Force between two blocks is 15 N.

Explanation:

1. Given data

  Mass of first block(m) = 2 kg

  Mass of second block (M) = 3 kg

  Coefficient of friction between first and second block $\mathbf{(\mu )=1}$

  Coefficient of friction between block and smooth horizontal  

  surface $\mathbf{(\mu )=0}$

  Force acting on first block (F) =25 N

2.Static friction force between first and second block

   $\mathbf{(F_{f})=\mu \times m\times g=1\times 2\times 10=20 \ N}$      ...1)

3. First consider both block together. Here is only one force(F) is acting.  

   There are no other force because friction force between block and surface is zero.

  From Newton' s second law

   $\mathbf{\sum F= Total \ mass\times acceleration}$

   25 = (2+3)×a

   25= 5×a

   So acceleration of whole system (a) = 5 $\mathbf{\frac{m}{s^{2}}}$     ...2)

4. Now consider first block which mass is 2 kg. On first block    

   one external force (25 N) and friction force is acting. Here    

   friction force is between first block and second block.  

  From Newton' s second law

   $\mathbf{\sum F= Total \ mass\times acceleration}$

  $\mathbf{F - Friction force =m\times a}$

  $\mathbf{25 - Friction force =2\times 5}$

  So

  Friction force = 15 N

  Friction force < $\mathbf{(F_{f})}$

  Friction force between first and second block is 15 N.    

  Means it do not reach the value of static friction force. It is  

  also clear that there are no slipping between first and  

  second block.