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Difficulty: EasyWork, Energy and Power

An object of mass 4 kg4\text{ kg} is released from rest at a height of 5 m5\text{ m} above the ground. Neglecting air resistance and taking g=10 m s−2g = 10\text{ m s}^{-2}, what is its kinetic energy just before striking the ground?

  1. 200 J200\text{ J}Answer
  2. B
    100 J100\text{ J}
  3. C
    40 J40\text{ J}
  4. D
    20 J20\text{ J}

Answer

The kinetic energy of the object just before striking the ground is 200 J200\text{ J}.
The total mechanical energy is conserved during free fall. The initial gravitational potential energy Ep=mgh=4×10×5=200 JE_p = mgh = 4 \times 10 \times 5 = 200\text{ J} is completely converted into kinetic energy EkE_k just before impact, making 200 J200\text{ J} the correct answer.

Step-by-Step Solution

1
Calculate the initial potential energy at maximum height
Ep=mgh=4 kg×10 m s−2×5 m=200 JE_p = mgh = 4\text{ kg} \times 10\text{ m s}^{-2} \times 5\text{ m} = 200\text{ J}
At the top of the fall, all mechanical energy is stored as gravitational potential energy.
2
Apply the principle of conservation of mechanical energy
Ek=Ep=200 JE_k = E_p = 200\text{ J}
In the absence of resistive forces such as air drag, potential energy lost converts completely into kinetic energy gained.

Key Concept

Conservation of Mechanical Energy
Estimated Time:45s
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Topics

Work, Energy and Power

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