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. 2J
B. 3.8 J
C. 5.2 J
D. 2.6 J

Answer

To find the work done by the force on a body of mass 3 kg, given the displacement equation \( S = \frac{t^3}{3} \) (in meters), we will follow these steps:

### Step 1: Differentiate the displacement to find velocity
The displacement \( S \) is given by:
\[ S = \frac{t^3}{3} \]
To find the velocity \( v \), we differentiate \( S \) with respect to time \( t \):
\[ v = \frac{dS}{dt} = \frac{d}{dt}\left(\frac{t^3}{3}\right) = t^2 \]

### Step 2: Differentiate the velocity to find acceleration
Next, we differentiate the velocity \( v \) to find the acceleration \( a \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(t^2) = 2t \]

### Step 3: Calculate the force using Newton's second law
Using Newton's second law, the force \( F \) can be calculated as:
\[ F = m \cdot a \]
Given that the mass \( m = 3 \, \text{kg} \):
\[ F = 3 \cdot (2t) = 6t \]

### Step 4: Set up the work done integral
The work done \( W \) by the force over a displacement can be expressed as:
\[ W = \int F \, ds \]
Since \( ds = v \, dt = t^2 \, dt \), we can substitute \( F \) and \( ds \):
\[ W = \int_0^2 (6t) \cdot (t^2) \, dt = \int_0^2 6t^3 \, dt \]

### Step 5: Evaluate the integral
Now we evaluate the integral:
\[ W = 6 \int_0^2 t^3 \, dt \]
The integral of \( t^3 \) is:
\[ \int t^3 \, dt = \frac{t^4}{4} \]
Thus,
\[ W = 6 \left[ \frac{t^4}{4} \right]_0^2 = 6 \left( \frac{2^4}{4} - \frac{0^4}{4} \right) = 6 \left( \frac{16}{4} \right) = 6 \cdot 4 = 24 \, \text{J} \]

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