question_answer

Two blocks of mass 2 kg and 3 kg are arranged as shown in the figure. The value of friction coefficient between 2 kg and 3 kg surface is \[0.4,\] and \[0.02t\] between the surface of 3 kg block and ground. A time varying horizontal external force \[F=5t\]is acting on 3 kg block (where t is time in sec.) Work done by the friction force on 2 kg block up to 5 sec with respect to 3 kg block is

A) \[100\text{ }J\]                 

B) \[400\text{ }J\]                 

C) zero          

D)  \[160\text{ }J\]

Correct Answer: C

Solution :

    [c] Maximum value of acceleration of 2 kg block is  \[\frac{0.4\times 20}{2}=4\,m/{{s}^{2}}\] Common acceleration \[=\frac{5t-0.02t\times 50}{5}=t-0.2t=0.8t\] So both blocks move together upto \[0.8t=4\Rightarrow 5\sec \] Up to 5 sec, static friction works between 2 kg and kg blocks. \[\therefore \] The work done by friction on 2 kg block w.r.t. 3 kg is zero.