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A ball tied to a rope making an angle `30^(@)` with the horizontal is held so that the rope is just If the ball is released, calculate the heat generated in joule when it crosses `30^(@)` down the horizontal (dotted)Given `m=Kg,L=2m,g=10m//s^(2)`. |
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Answer» The speed just after jerking `v=sqrt(2gL) cos 30^(@)=sqrt((3gL)/(2))` `therefore` heat=10 sec in kinetic energy `=(1)/(2)m(2gL-(3gL)/(2))` `=(mgL)/(4)=5J` |
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