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The 30.0-kg package above starts at 0.500 m/s and receives 92.0 J of net work while being pushed 0.800 m. Using the work-energy theorem, what is its final speed?
A3.07 m/s
B1.94 m/s
C2.53 m/s
D6.14 m/s
Answer & Solution
Correct answer: C. 2.53 m/s
1. The work-energy theorem gives one half m v^2 = Wnet + one half m v0^2.
2. The initial kinetic energy is one half (30.0 kg)(0.500 m/s)^2 = 3.75 J.
3. Adding the net work, one half m v^2 = 92.0 J + 3.75 J = 95.75 J.
4. Solving for v gives v = the square root of (2 times 95.75 J divided by 30.0 kg), which is the square root of 6.383 m^2/s^2.
5. Taking that square root gives v = 2.53 m/s.
6. 1.94 m/s and 3.07 m/s come from arithmetic slips in the division or square root, and 6.14 m/s comes from forgetting to take the square root at all.
_Source: OpenStax College Physics (CC BY 4.0), Ch 7 "Work, Energy, and Energy Resources", section 7.2 Kinetic Energy and the Work-Energy Theorem_
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