A body of mass 4kg 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 SI system as

Options

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

Explanation

To solve the problem of calculating the work done by the force on a body of mass 4 kg, given its position as a function of time, we can follow these steps:

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

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

### Step 3: Calculate the velocity at t = 3 seconds
Now, we can find the velocity at \( t = 3 \) seconds:
\[ v(3) = (3)^3 = 27 \, \text{m/s} \]

### Step 4: Calculate the initial velocity
At \( t = 0 \):
\[ v(0) = (0)^3 = 0 \, \text{m/s} \]

### Step 5: Calculate the change in kinetic energy
The work done by the force is equal to the change in kinetic energy. The kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} mv^2 \]
The change in kinetic energy from \( t = 0 \) to \( t = 3 \) seconds is:
\[ \Delta KE = KE(3) - KE(0) \]
Calculating \( KE(3) \):
\[ KE(3) = \frac{1}{2} \times 4 \times (27)^2 \]
\[ KE(3) = 2 \times 729 = 1458 \, \text{J} \]

At \( t = 0 \):
\[ KE(0) = \frac{1}{2} \times 4 \times (0)^2 = 0 \, \text{J} \]

Thus, the change in kinetic energy is:
\[ \Delta KE = 1458 - 0 = 1458 \, \text{J} \]

### Step 6: Conclusion
The work done by the force in the first three seconds is:
\[ W = \Delta KE = 1458 \, \text{J} \]

### Final Answer
The work done by the force in the first three seconds is **1458 Joules**.

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