A body of mass 20 kg is dropped from a height of 2 m. If g is taken to be equal to 10 m/`s^(2)`, the kinetic energy of the body, just before striking the ground, will be

Options

A. 400 J
B. 4 J
C. 40 J
D. none of these

Explanation

To find the kinetic energy of a body just before it strikes the ground, we can use the principle of conservation of energy. The potential energy at the height will be converted into kinetic energy just before it hits the ground.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the body (m) = 20 kg
- Height (h) = 2 m
- Acceleration due to gravity (g) = 10 m/s²

2. **Calculate the potential energy (PE) at the height:**
The potential energy at height h is given by the formula:
\[ PE = mgh \]
Substituting the values:
\[ PE = 20 \, \text{kg} \times 10 \, \text{m/s}^2 \times 2 \, \text{m} = 400 \, \text{J} \]

3. **Apply the conservation of energy principle:**
According to the conservation of energy, the total mechanical energy at the height (potential energy) will be equal to the kinetic energy (KE) just before hitting the ground:
\[ KE = PE \]

4. **Conclusion:**
Therefore, the kinetic energy of the body just before striking the ground is:
\[ KE = 400 \, \text{J} \]

### Final Answer:
The kinetic energy of the body just before striking the ground is **400 Joules**.

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