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Grade: 9
        

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

Answers : (1)

Arun
22989 Points
							

According to law of conservation of mechanical energy,

Initial mech. energy = final mech. Energy

Or,. 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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