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Difficulty: MediumWork, Energy, and Power

A maintenance technician drops a 4-kilogram4\text{-kilogram} mechanical component from an elevated platform located 5 meters5\text{ meters} above the workshop floor. Assuming air resistance is negligible and the acceleration due to gravity is 9.8 m/s29.8\text{ m/s}^2, how much kinetic energy does the component possess immediately before striking the floor?

  1. A
    0 J0\text{ J}
  2. B
    39.2 J39.2\text{ J}
  3. 196 J196\text{ J}Answer
  4. D
    20 J20\text{ J}

Answer

The kinetic energy immediately before impact is 196 J196\text{ J}.
According to the principle of conservation of energy, the mechanical energy in an isolated system remains constant. As the component falls, its initial gravitational potential energy (PE=mghPE = mgh) is entirely converted into kinetic energy (KEKE) right before impact: KE=4 kg×9.8 m/s2×5 m=196 JKE = 4\text{ kg} \times 9.8\text{ m/s}^2 \times 5\text{ m} = 196\text{ J}.

Step-by-Step Solution

1
Calculate the initial gravitational potential energy at the release height.
PE=mgh=4 kg×9.8 m/s2×5 m=196 JPE = mgh = 4\text{ kg} \times 9.8\text{ m/s}^2 \times 5\text{ m} = 196\text{ J}
At the moment of release from rest, all mechanical energy of the component is stored as gravitational potential energy.
2
Apply the law of conservation of mechanical energy.
KEfinal=PEinitial=196 JKE_{\text{final}} = PE_{\text{initial}} = 196\text{ J}
In the absence of non-conservative forces like air resistance, total mechanical energy is conserved, meaning all potential energy converts into kinetic energy at ground level (h=0h = 0).

Key Concept

Conservation of Mechanical Energy
Estimated Time:1m 0s
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Topics

Work, Energy, and Power

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