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How much work does it take to accelerate a 9.0 kg object from rest to 27 m/s? Group of answer choices 2200 J
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Answers ( )
initial kinetic energy = 0J (v=0 (rest))
Final kinetic energy = 1/2mv ²
Ek=1/2(9)(27 ²)
Ek=4.5(729)
Ek=3280.5
∆Energy=3280.5J
(As it starts from 0)
Work= ∆energy
So work =3280.5J
Answer=3300J (forth option)