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 Kinetic Energy (KE_initial):**
- The object falls from rest, so the initial velocity \( u = 0 \).
- Therefore, the initial kinetic energy is:
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 = \frac{1}{2} \times 10 \times 0^2 = 0 \, \text{J} \]

3. **Calculate the Final Kinetic Energy (KE_final):**
- The final kinetic energy when the object reaches a velocity of \( 10 \, \text{m/s} \) is:
\[ KE_{\text{final}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 10 \times (10)^2 = \frac{1}{2} \times 10 \times 100 = 500 \, \text{J} \]

4. **Calculate the Work Done by Gravity (W_gravity):**
- The work done by gravity when the object falls through a height of \( 10 \, \text{m} \) is:
\[ W_{\text{gravity}} = mgh = 10 \times 10 \times 10 = 1000 \, \text{J} \]

5. **Apply the Work-Energy Theorem:**
- According to the work-energy theorem:
\[ W_{\text{total}} = KE_{\text{final}} - KE_{\text{initial}} \]
- Thus, the total work done on the object is:
\[ W_{\text{total}} = 500 - 0 = 500 \, \text{J} \]

6. **Calculate the Work Done by Air Resistance (W_air):**
- The total work done is the sum of the work done by gravity and the work done by air resistance:
\[ W_{\text{total}} = W_{\text{gravity}} + W_{\text{air}} \]
- Rearranging gives:
\[ W_{\text{air}} = W_{\text{total}} - W_{\text{gravity}} = 500 - 1000 = -500 \, \text{J} \]

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

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