A 30.0-kg package on a roller belt conveyor moves at 0.500 m/s. What is its kinetic energy?
- 15.0 J
- 7.50 J
- 3.75 J
- 0.500 J
Show answer and solution
Correct answer: C. 3.75 J
1. Kinetic energy is KE = one half m v squared.
2. Substituting the given values, KE = (0.5)(30.0 kg)(0.500 m/s)^2.
3. Squaring the speed gives 0.250 m^2/s^2, and multiplying by 0.5 times 30.0 kg gives KE = 3.75 J.
4. 15.0 J and 7.50 J come from forgetting to square the speed or forgetting the factor of one half, and 0.500 J is simply the speed left as if it were the answer.
_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_