A body of mass 2 kg fall vertically, passing through two points A and B. The speeds of the body as it passes A and B are 1 m/s and 4m/s respectively. The resistance against which the body falls is 9.6N. What is the distance AB?

Options

A. 2m
B. 3m
C. 6m
D. 1.5 m

Answer

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

### Step-by-Step Solution:

1. **Identify the Known Values**:
- Mass of the body, \( m = 2 \, \text{kg} \)
- Speed at point A, \( v_A = 1 \, \text{m/s} \)
- Speed at point B, \( v_B = 4 \, \text{m/s} \)
- Resistance force, \( F_{\text{resistance}} = 9.6 \, \text{N} \)
- Gravitational acceleration, \( g \approx 9.8 \, \text{m/s}^2 \)

2. **Calculate the Change in Kinetic Energy**:
\[ KE_A = \frac{1}{2} m v_A^2 = \frac{1}{2} \times 2 \times (1)^2 = 1 \, \text{J} \]
\[ KE_B = \frac{1}{2} m v_B^2 = \frac{1}{2} \times 2 \times (4)^2 = 16 \, \text{J} \]
\[ \Delta KE = KE_B - KE_A = 16 \, \text{J} - 1 \, \text{J} = 15 \, \text{J} \]

3. **Calculate the Work Done by Forces**:
The work done by the gravitational force and the resistance force can be expressed as:
\[ W = F_{\text{gravity}} \cdot h - F_{\text{resistance}} \cdot h \]
Where \( F_{\text{gravity}} = m \cdot g = 2 \cdot 9.8 = 19.6 \, \text{N} \).

Thus, the work done can be written as:
\[ W = (19.6 \, \text{N} - 9.6 \, \text{N}) \cdot h = 10 \cdot h \]

4. **Set Work Done Equal to Change in Kinetic Energy**:
According to the work-energy theorem:
\[ W = \Delta KE \]
Therefore:
\[ 10h = 15 \]

5. **Solve for Distance \( h \)**:
\[ h = \frac{15}{10} = 1.5 \, \text{m} \]

### Final Answer:
The distance \( AB \) is \( 1.5 \, \text{m} \).

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