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3. Cupid shot an arrow with a mass of .63kg. The arrow flies at the unknowing target with a speed of 432m/s. a) What is the kinetic energy of an arrow? Givens: Formula: Sub: Solve: b) What is the Work done?
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a) Kinetic energy:
Formula: KE = 1/2 m v^2
Substitute: KE = 0.5(0.63 kg)(432 m/s)^2
Solve: (432)^2 = 186,624 → KE = 0.315 × 186,624 = 58,786.56 J
So KE ≈ 5.88 × 10^4 J (≈ 58.8 kJ).
b) Work done:
The work done to accelerate the arrow from rest equals the change in kinetic energy, W = ΔKE = KEfinal − KEinitial = KE − 0 = 58,786.56 J.
So W ≈ 5.88 × 10^4 J.
Formula: KE = 1/2 m v^2
Substitute: KE = 0.5(0.63 kg)(432 m/s)^2
Solve: (432)^2 = 186,624 → KE = 0.315 × 186,624 = 58,786.56 J
So KE ≈ 5.88 × 10^4 J (≈ 58.8 kJ).
b) Work done:
The work done to accelerate the arrow from rest equals the change in kinetic energy, W = ΔKE = KEfinal − KEinitial = KE − 0 = 58,786.56 J.
So W ≈ 5.88 × 10^4 J.
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