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`        A body of mass m dropped from a height H reaches the ground with a speed of 1.2gh1/2. Then thework done by air friction is :-`
one year ago

Arun
23353 Points
```							According to law of conservation of mechanical energy,Initial mech. energy = final mech. EnergyOr,. K(initial) + U(initial)= K(final)+ U (final)Since ,initially(at H height) the body only have potential energy not kintetic energy. Hence at this point the “K (initial )” becomes ‘zero’.But, finally (at the bottom) ,it has only kintetic energy. Hence at this point the potential energy, “U (final)” becomes 'zero'.Therefore, U (initial) = K ( final).This condition only exist when there is only conservative force,. But here friction force is present ,which is non conservative force , so here some energy becomes lost due to friction ,which is equals to work done by friction .so the above relation becomes--K(final)-U (initial)= loss=1/2mv^2- mgH=1/2m(1.2√gH)^2- mgH=1/2m(1.44gH)-mgH now you can solve it
```
one year ago
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• Test paper with Video Solution
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• NCERT Solutions
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• Previous Year Exam Questions