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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
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
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
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