Under the action of a force, a `2 kg` body moves such that its position x as a function of time is given by `x =(t^(3))/(3)` where x is in meter and t in second. The work done by the force in the first two seconds is .

Options

A. `1600 J`
B. `160 J`
C. `16 J`
D. `1.6 J`

Explanation

To solve the problem step by step, we will calculate the work done by the force acting on a 2 kg body whose position as a function of time is given by \( x = \frac{t^3}{3} \).

### Step 1: Determine the velocity of the body
The velocity \( v \) can be found by differentiating the position function \( x(t) \) with respect to time \( t \).

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^3}{3}\right) = \frac{3t^2}{3} = t^2 \]

### Step 2: Calculate the kinetic energy at \( t = 0 \) seconds
The kinetic energy \( KE \) is given by the formula:

\[ KE = \frac{1}{2}mv^2 \]

At \( t = 0 \):

\[ v(0) = 0^2 = 0 \]

Thus, the kinetic energy at \( t = 0 \) is:

\[ KE(0) = \frac{1}{2} \times 2 \, \text{kg} \times (0)^2 = 0 \, \text{Joules} \]

### Step 3: Calculate the kinetic energy at \( t = 2 \) seconds
Now, we calculate the velocity at \( t = 2 \) seconds:

\[ v(2) = 2^2 = 4 \, \text{m/s} \]

Now, we can find the kinetic energy at \( t = 2 \):

\[ KE(2) = \frac{1}{2} \times 2 \, \text{kg} \times (4)^2 = \frac{1}{2} \times 2 \times 16 = 16 \, \text{Joules} \]

### Step 4: Calculate the work done by the force
The work done \( W \) by the force is the change in kinetic energy from \( t = 0 \) to \( t = 2 \):

\[ W = KE(2) - KE(0) = 16 \, \text{Joules} - 0 \, \text{Joules} = 16 \, \text{Joules} \]

### Final Answer
The work done by the force in the first two seconds is \( \boxed{16 \, \text{Joules}} \).

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