A body of mass 4 kg is accelerated up by a constant force, travels a distance of 5 m in the first second and a distance of 2m in the third second. The force acting on the body is

Options

A. 2N
B. 4N
C. 6N
D. 8N

Explanation

To solve the problem, we need to determine the force acting on a body of mass 4 kg that travels different distances in the first and third seconds under constant acceleration.

### Step-by-step Solution:

1. **Identify Given Data**:
- Mass of the body (m) = 4 kg
- Distance traveled in the first second (s1) = 5 m
- Distance traveled in the third second (s3) = 2 m

2. **Use the Equation of Motion**:
The distance traveled in the nth second can be expressed using the formula:
\[ s_n = u + \frac{1}{2} a (2n - 1) \]
where:
- \(s_n\) = distance traveled in the nth second
- \(u\) = initial velocity
- \(a\) = acceleration
- \(n\) = time in seconds

3. **Set Up Equations**:
- For the first second (n = 1):
\[ s_1 = u + \frac{1}{2} a (2 \cdot 1 - 1) = u + \frac{1}{2} a \]
Thus, we have:
\[ 5 = u + \frac{1}{2} a \quad \text{(Equation 1)} \]

- For the third second (n = 3):
\[ s_3 = u + \frac{1}{2} a (2 \cdot 3 - 1) = u + \frac{1}{2} a \cdot 5 \]
Thus, we have:
\[ 2 = u + \frac{5}{2} a \quad \text{(Equation 2)} \]

4. **Solve the Equations**:
- From Equation 1:
\[ u = 5 - \frac{1}{2} a \quad \text{(Substitute this into Equation 2)} \]

- Substitute \(u\) into Equation 2:
\[ 2 = (5 - \frac{1}{2} a) + \frac{5}{2} a \]
Simplifying this gives:
\[ 2 = 5 - \frac{1}{2} a + \frac{5}{2} a \]
\[ 2 = 5 + 2a \]
\[ 2a = 2 - 5 \]
\[ 2a = -3 \]
\[ a = -\frac{3}{2} \, \text{m/s}^2 \]

5. **Calculate the Force**:
Using Newton's second law:
\[ F = m \cdot a \]
Substitute the values:
\[ F = 4 \cdot \left(-\frac{3}{2}\right) = -6 \, \text{N} \]

The negative sign indicates that the force is acting in the opposite direction to the acceleration.

### Final Answer:
The magnitude of the force acting on the body is **6 N**.