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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

6600 J

120 J

3300 J

0.33 J

Sagot :

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)
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