A block 'A' of mass 'm' is attached at one end of a light spring and the other end of the spring is connected to another block 'B' of mass 2 m through a light string as shown in the figure.'A' is held and B is in static equilibrium. Now A is released. The acceleration of A just after that instant is 'a'. In the next case, B is held and A is in static equilibrium. Now when,B is released, its acceleration immediately after the release is 'b'. The value of a/b is: (Pulley, string and the spring are massless)

Question

Infinity Learn · Accepted Answer

For first case tension in spring will beTs $$=$$ $$2mg$$ just after 'A' is $$released.2mg-mg$$ $$=$$ ma ⇒ a $$=$$ gIn second case Ts $$=$$ mg $$2mg$$ $$-$$ mg $$=$$ $$2mbb$$ $$=$$ $$g2ab=$$ $$2$$

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material