An object of mass 10 kg falls from rest through a vertical distance of 10m and acquires a velocity of 10m/s . The work done by the push of air on the object is (g=10 m/s 2 )

a) 500J b)-500J c)250J d)-250J

The forces acting on the object are:

1-Weight of object mg (downward)
2-Push of air F(Acting upward)
The net force in the downward direction=mg-F
Work in falling 10 m=Gain in K.E
=(mg-F) x 10=1/2 mv 2
mg-F=1/20 mv 2
F=mg-1/20mv 2=10x10-0.05x10x100=50J
Work done by the push of air=FScos180=50 x10xcos 180=-500J

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