A body of mass 4 kg is acted upon by a force. The position of body with respect to time is denoted by `x=(t^4)/(4)`. The work done by the force in first three seconds can be expressed in (J)

Options

A. 1458
B. 729
C. 54
D. 27

Explanation

To solve the problem step by step, we will follow the work-energy theorem, which states that the work done by the force on an object is equal to the change in its kinetic energy.

### Step 1: Determine the position function
The position of the body with respect to time is given by:
\[ x(t) = \frac{t^4}{4} \]

### Step 2: Find the velocity
To find the velocity, we differentiate the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^4}{4}\right) = t^3 \]

### Step 3: Find the kinetic energy
The kinetic energy \( K \) of the body is given by the formula:
\[ K = \frac{1}{2}mv^2 \]
where \( m \) is the mass of the body and \( v \) is its velocity.

Given that the mass \( m = 4 \, \text{kg} \), we can substitute the expression for velocity:
\[ K(t) = \frac{1}{2} \cdot 4 \cdot (t^3)^2 = \frac{1}{2} \cdot 4 \cdot t^6 = 2t^6 \]

### Step 4: Calculate the initial kinetic energy
At \( t = 0 \):
\[ K_{\text{initial}} = K(0) = 2(0)^6 = 0 \, \text{J} \]

### Step 5: Calculate the final kinetic energy at \( t = 3 \, \text{s} \)
At \( t = 3 \):
\[ K_{\text{final}} = K(3) = 2(3)^6 \]
Calculating \( 3^6 \):
\[ 3^6 = 729 \]
Thus,
\[ K_{\text{final}} = 2 \cdot 729 = 1458 \, \text{J} \]

### Step 6: Calculate the work done
According to the work-energy theorem:
\[ \text{Work done} = K_{\text{final}} - K_{\text{initial}} \]
Substituting the values we found:
\[ \text{Work done} = 1458 \, \text{J} - 0 \, \text{J} = 1458 \, \text{J} \]

### Final Answer
The work done by the force in the first three seconds is:
\[ \boxed{1458 \, \text{J}} \]