Under the action of foece, 1 kg body moves such that its position x as a function of time t is given by `x=(t^(3))/(3),` x is meter. Calculate the work done (in joules) by the force in first 2 seconds.

Answer

To solve the problem of calculating the work done by the force on a 1 kg body moving under the given position function \( x(t) = \frac{t^3}{3} \) for the first 2 seconds, we can follow these steps:

### Step 1: Determine the Force Acting on the Body
The force \( F \) acting on the body can be calculated using Newton's second law, which states that \( F = ma \), where \( m \) is the mass and \( a \) is the acceleration.

Given:
- Mass \( m = 1 \, \text{kg} \)

### Step 2: Calculate the Acceleration
To find the acceleration, we need to differentiate the position function \( x(t) \) twice with respect to time \( t \).

1. **First Derivative (Velocity)**:
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^3}{3}\right) = t^2 \]

2. **Second Derivative (Acceleration)**:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(t^2) = 2t \]

### Step 3: Calculate the Force
Now we can calculate the force:
\[ F = ma = 1 \cdot (2t) = 2t \, \text{N} \]

### Step 4: Calculate the Work Done
The work done \( W \) by the force over a displacement can be calculated using the integral:
\[ W = \int F \, dx \]

Since \( dx = v(t) \, dt = t^2 \, dt \), we can substitute \( F \) and \( dx \) into the integral:
\[ W = \int_0^2 F \, dx = \int_0^2 (2t)(t^2) \, dt = \int_0^2 2t^3 \, dt \]

### Step 5: Evaluate the Integral
Now we evaluate the integral:
\[ W = 2 \int_0^2 t^3 \, dt = 2 \left[\frac{t^4}{4}\right]_0^2 \]
Calculating the limits:
\[ = 2 \left[\frac{2^4}{4} - \frac{0^4}{4}\right] = 2 \left[\frac{16}{4}\right] = 2 \cdot 4 = 8 \, \text{J} \]

### Final Result
Thus, the work done by the force in the first 2 seconds is:
\[ W = 8 \, \text{J} \]