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