An object of mass 10 kg falls from rest through a vertical distance of 10 m and acquires a velocity of `10 ms^(-1)`. The work done by the push of air on the object is `(g = 10 ms^(-2)`)

Options

A. 500 J
B. `-500 J`
C. `250 J`
D. `-250 J`

Answer

To solve the problem, we will use the work-energy theorem, which states that the work done by all forces acting on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the object, \( m = 10 \, \text{kg} \)
- Distance fallen, \( h = 10 \, \text{m} \)
- Final velocity, \( v = 10 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Initial and Final Kinetic Energy:**
- Initial kinetic energy, \( KE_i = 0 \) (since the object starts from rest)
- Final kinetic energy, \( KE_f = \frac{1}{2} m v^2 = \frac{1}{2} \times 10 \times (10)^2 = \frac{1}{2} \times 10 \times 100 = 500 \, \text{J} \)

3. **Calculate the Work Done by Gravity:**
- Work done by gravity, \( W_g = mgh = 10 \times 10 \times 10 = 1000 \, \text{J} \)

4. **Apply the Work-Energy Theorem:**
- According to the work-energy theorem:
\[ W_{\text{total}} = KE_f - KE_i \]
- Here, \( W_{\text{total}} = W_g + W_{\text{air}} \)
- Therefore, we can write:
\[ W_g + W_{\text{air}} = KE_f - KE_i \]
- Substituting the known values:
\[ 1000 + W_{\text{air}} = 500 - 0 \]
- This simplifies to:
\[ W_{\text{air}} = 500 - 1000 = -500 \, \text{J} \]

5. **Conclusion:**
- The work done by the push of air on the object is \( W_{\text{air}} = -500 \, \text{J} \).

### Final Answer:
The work done by the push of air on the object is \( -500 \, \text{J} \).

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