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 find the work done by the force on a 2 kg body moving under the given position function \( x(t) = \frac{t^3}{3} \) in the first two seconds, we can follow these steps:

### Step 1: Determine the velocity function
The velocity \( v(t) \) is the derivative of the position function \( x(t) \) with respect to time \( t \).

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

### Step 2: Calculate the initial and final velocities
Now, we need to find the velocities at \( t = 0 \) seconds and \( t = 2 \) seconds.

- At \( t = 0 \):
\[ v(0) = 0^2 = 0 \, \text{m/s} \]

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

### Step 3: Calculate the change in kinetic energy
The work done by the force is equal to the change in kinetic energy (KE) of the body. The kinetic energy is given by the formula:

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

Now we can calculate the initial and final kinetic energies.

- Initial kinetic energy \( KE_0 \) at \( t = 0 \):
\[ KE_0 = \frac{1}{2} \times 2 \, \text{kg} \times (0 \, \text{m/s})^2 = 0 \, \text{J} \]

- Final kinetic energy \( KE_2 \) at \( t = 2 \):
\[ KE_2 = \frac{1}{2} \times 2 \, \text{kg} \times (4 \, \text{m/s})^2 = \frac{1}{2} \times 2 \times 16 = 16 \, \text{J} \]

### Step 4: Calculate the work done
The work done \( W \) is the change in kinetic energy:

\[ W = KE_2 - KE_0 = 16 \, \text{J} - 0 \, \text{J} = 16 \, \text{J} \]

Thus, the work done by the force in the first two seconds is \( \boxed{16 \, \text{J}} \).

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