A body of mass `3 kg` is under a constant force which causes a displacement `s` metre in it, given by the relation `s=(1)/(3)t^(2)`, where `t` is in seconds. Work done by the force in 2 seconds is

Options

A. `(5)/(19)J`
B. `(3)/(8)J`
C. `(8)/(3)J`
D. `(19)/(5)J`

Answer

To solve the problem step by step, we will use the given information about the displacement of the body and apply the work-energy theorem.

### Step 1: Understand the given displacement equation
The displacement \( s \) of the body is given by the equation:
\[ s = \frac{1}{3} t^2 \]
where \( t \) is the time in seconds.

### Step 2: Find the velocity of the body
Velocity \( v \) is the derivative of displacement with respect to time. Therefore, we differentiate \( s \) with respect to \( t \):
\[ v = \frac{ds}{dt} = \frac{d}{dt}\left(\frac{1}{3} t^2\right) = \frac{2}{3} t \]

### Step 3: Calculate the initial and final velocities
- At \( t = 0 \) seconds:
\[ v_1 = \frac{2}{3} \cdot 0 = 0 \, \text{m/s} \]
- At \( t = 2 \) seconds:
\[ v_2 = \frac{2}{3} \cdot 2 = \frac{4}{3} \, \text{m/s} \]

### Step 4: Use the work-energy theorem
The work done \( W \) by the force is equal to the change in kinetic energy:
\[ W = \Delta KE = \frac{1}{2} m (v_2^2 - v_1^2) \]
Given that the mass \( m = 3 \) kg, we can substitute \( v_1 \) and \( v_2 \):
\[ W = \frac{1}{2} \cdot 3 \left(\left(\frac{4}{3}\right)^2 - 0^2\right) \]

### Step 5: Calculate \( v_2^2 \)
\[ v_2^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]

### Step 6: Substitute and simplify the work done
Now substituting \( v_2^2 \) into the work equation:
\[ W = \frac{1}{2} \cdot 3 \cdot \frac{16}{9} = \frac{3 \cdot 16}{2 \cdot 9} = \frac{48}{18} = \frac{8}{3} \, \text{Joules} \]

### Final Answer
The work done by the force in 2 seconds is:
\[ \boxed{\frac{8}{3} \, \text{Joules}} \]