A body of mass `3 kg` is under a force , which causes a displacement in it is given by `S = (t^(3))/(3)` (in metres). Find the work done by the force in first `2` seconds.

Options

A. `2 J`
B. `3.8 J`
C. `5.2 J`
D. `24 J`

Explanation

To solve the problem of finding the work done by a force on a body of mass 3 kg that undergoes a displacement given by \( S = \frac{t^3}{3} \) over the first 2 seconds, we can follow these steps:

### Step 1: Understand the displacement function
The displacement \( S \) is given by:
\[ S = \frac{t^3}{3} \]
This function describes how the position of the body changes with time.

### Step 2: Find the velocity
Velocity \( v \) is the rate of change of displacement with respect to time. We can find it by differentiating \( S \) with respect to \( t \):
\[ v = \frac{dS}{dt} = \frac{d}{dt}\left(\frac{t^3}{3}\right) = t^2 \]

### Step 3: Find the acceleration
Acceleration \( a \) is the rate of change of velocity with respect to time. We can find it by differentiating \( v \) with respect to \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(t^2) = 2t \]

### Step 4: Apply Newton's second law
According to Newton's second law, the force \( F \) acting on the body can be expressed as:
\[ F = ma \]
where \( m \) is the mass of the body. Given that \( m = 3 \, \text{kg} \) and \( a = 2t \), we have:
\[ F = 3 \cdot (2t) = 6t \]

### Step 5: Calculate the work done
The work done \( W \) by the force over a displacement \( dS \) can be expressed as:
\[ dW = F \cdot dS \]
We need to integrate this expression over the time interval from \( 0 \) to \( 2 \) seconds. Since \( dS = v \, dt = t^2 \, dt \), we can substitute \( F \) and \( dS \) into the work done formula:
\[ W = \int_0^2 F \, dS = \int_0^2 (6t) (t^2 \, dt) \]
This simplifies to:
\[ W = \int_0^2 6t^3 \, dt \]

### Step 6: Evaluate the integral
Now we evaluate the integral:
\[ W = 6 \int_0^2 t^3 \, dt = 6 \left[ \frac{t^4}{4} \right]_0^2 = 6 \left[ \frac{2^4}{4} - \frac{0^4}{4} \right] \]
Calculating this gives:
\[ W = 6 \left[ \frac{16}{4} \right] = 6 \cdot 4 = 24 \, \text{Joules} \]

### Final Answer
The work done by the force in the first 2 seconds is:
\[ \boxed{24 \, \text{Joules}} \]