the bob of a pendulum is released from horizontal posn a.if the lengthof pendulum is 1.5 m what is the speed with which the bob arrives at lowermost point b, given that it dissipated 5% of its energy against air resistance?

Dear manisha

initial energy of bob = mg*1.5

energy dissipated = 5/100 * (mg*1.5)

now energy balane

 mg*1.5 = 1/2 m v² + 5/100 * (mg*1.5)

 v² =3g - 3g/20

v² =57g/20

v= √(57g/20)

   
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