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akshay singh Grade: 12
        THE BOB OF A PENDULUM IS RELEASED FROM HORIZONTAL POSITION.IF THE LENGTH OF PENDULUM IS 1.5 M,WHAT IS SPEED WITH WHICH THE BOB COME AT LOWEST POINT.GIVEN THAT IT DISSIPATES 5% OF ITS INITIAL ENERGY AGAINST AIR RESISTANCE
8 years ago

Answers : (1)

askIITIians Expert
21 Points
										

 



 


Let the reference level as given. Here the potential energy be zero.


From energy conservation


Potential Energy at B = K.E at A


Mgh = 1/2 MV2          ......................................... Where V is the velocity at A


Mgh - 5% Mgh = 1/2 MV2          ....................... Given


gh - 5% gh = 1/2 V2          .................................... (As 5% of energy is lost against air resistance)


gh - 5/100gh = 1/2 V2


95/100 gh = 1/2 V2


V2 = 190/100 gh



Thus bob comes down with speed of 5.33 m/s

8 years ago
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