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Q5.34  Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (b) the action-reaction forces between A and B ? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between \mu_s and \mu_k.

    

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Consider block A.

      The frictional force on block A will be :

                                 f_A\ =\ \mu m_Ag

or                                      =\ 0.15\times5\times 10\ =\ 7.5\ N

Thus net force on B due to block A is  =  200  -  7.5   = 192.5 N.

The net force on the partition is 177.5 N.

Using Newton's law of motion we have, 

                                  a\ =\ \frac{Force}{m_A\ +\ m_B}

or                               a\ =\ \frac{177.5}{5+10}\ =\ 11.83\ m/s^2

For block A :

                                   F_A\ =\ m_Aa

or                                         =\ 5\times 11.83\ =\ 59.15\ N

So the required normal force is   =    192.15  -  59.15  =  133.35  N.                                       

Posted by

Devendra Khairwa

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