An object of mass 26 kg is thrown straight upwards from the ground at 27 m/s. Due to air resistance (a non-conservative force), the object loses some of its initial energy during its trip to reach maximum height and only travels to a height above the ground of 16.3 meters. What percentage of its initial energy is lost to air resistance during its flight to maximum height?

asked by Chelsea

12 years ago

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1 answer

Without air resistance
KE=PE
mv²/2=mgh
h=v²/2g =27²/2•9.8 = 37.2 m
(mgh-mgh1)/mgh = (h-h1)/h= (37.2-16.3)/37.2=0.56
56%

answered by Elena

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Question

An object of mass 26 kg is thrown straight upwards from the ground at 27 m/s. Due to air resistance (a non-conservative force), the object loses some of its initial energy during its trip to reach maximum height and only travels to a height above the ground of 16.3 meters. What percentage of its initial energy is lost to air resistance during its flight to maximum height?

1 answer

Without air resistance
KE=PE
mv²/2=mgh
h=v²/2g =27²/2•9.8 = 37.2 m
(mgh-mgh1)/mgh = (h-h1)/h= (37.2-16.3)/37.2=0.56
56%

Elena · Answer

Without air resistance KE=PE mv²/2=mgh h=v²/2g =27²/2•9.8 = 37.2 m (mgh-mgh1)/mgh = (h-h1)/h= (37.2-16.3)/37.2=0.56 56%