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The blocks A and B shown in the figure have masses MA = 5 kg and MB = 4 kg. The system is released from rest. The speed of B after A has travelled a distance 1 m along the incline is 

 

Option: 1

\frac{\sqrt{3}}{2}\sqrt{g}


Option: 2

\frac{\sqrt{3}}{4}\sqrt{g}


Option: 3

\frac{\sqrt{g}}{2\sqrt{3}}


Option: 4

\frac{\sqrt{g}}{2}


Answers (1)

best_answer

        

Work Energy Theorem -

Work done by all the forces acting on a particle equal to change in kinetic energy

-

 

 

   If A moves down the incline by 1 metre, B shall move up by \frac{1}{2} \ m metre. If the speed of B is v then the speed of A will be 2v.

          From conservation of energy:

            Gain in K.E. = loss in P.E.

\frac{1}{2}m_{A}(2v)^{2}+\frac{1}{2}m_{B}v^{2}=m_{A}g \times \frac{3}{5}-m_{B}g \times \frac{1}{2}

Solving we get

v=\frac{1}{2}\sqrt{\frac{g}{3}}

Posted by

Divya Prakash Singh

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