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. 1.5m

Answer

To solve the problem, we will use the principles of work, energy, and the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Mass of the body (m) = 2 kg
- Initial speed at point A (u) = 1 m/s
- Final speed at point B (v) = 4 m/s
- Resistance force (F_resistance) = 9.6 N

### Step 2: Calculate the net force acting on the body
The net force (F_net) acting on the body can be calculated using Newton's second law. The weight of the body (W) is given by:
\[ W = m \cdot g \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

So,
\[ W = 2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \]

Now, the net force acting on the body is:
\[ F_{net} = W - F_{resistance} = 19.6 \, \text{N} - 9.6 \, \text{N} = 10 \, \text{N} \]

### Step 3: Calculate the acceleration of the body
Using Newton's second law:
\[ F_{net} = m \cdot a \]
we can rearrange to find acceleration (a):
\[ a = \frac{F_{net}}{m} = \frac{10 \, \text{N}}{2 \, \text{kg}} = 5 \, \text{m/s}^2 \]

### Step 4: Use the equation of motion to find the distance AB
We can use the equation of motion:
\[ v^2 - u^2 = 2as \]
where:
- \( v \) = final velocity = 4 m/s
- \( u \) = initial velocity = 1 m/s
- \( a \) = acceleration = 5 m/s²
- \( s \) = distance AB

Substituting the values:
\[ (4 \, \text{m/s})^2 - (1 \, \text{m/s})^2 = 2 \cdot 5 \, \text{m/s}^2 \cdot s \]
\[ 16 - 1 = 10s \]
\[ 15 = 10s \]
\[ s = \frac{15}{10} = 1.5 \, \text{m} \]

### Conclusion
The distance AB is \( 1.5 \, \text{m} \).

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