Displacement x ( in meters ) , of a body of mass 1 kg as a function of time t, on a horizontal smooth surface , is given as `x = 2 t^2` Then work done in the first one second by the external force is

Options

A. 2 J
B. 4 J
C. 8 J
D. 16 J

Answer

To solve the problem, we need to find the work done by the external force on a body of mass 1 kg, given that its displacement \( x \) as a function of time \( t \) is given by the equation:

\[ x = 2t^2 \]

### Step-by-Step Solution:

1. **Determine the Velocity**:
The velocity \( v \) can be found by differentiating the displacement \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d(2t^2)}{dt} = 4t \]

2. **Determine the Acceleration**:
The acceleration \( a \) is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} = \frac{d(4t)}{dt} = 4 \, \text{m/s}^2 \]

3. **Calculate the Force**:
Using Newton's second law, \( F = ma \):
\[ F = 1 \, \text{kg} \times 4 \, \text{m/s}^2 = 4 \, \text{N} \]

4. **Calculate the Work Done**:
Work done \( W \) by the force over the first second can be calculated using the formula:
\[ W = \int_{0}^{1} F \, dx \]
Since \( F \) is constant (4 N), we can factor it out of the integral:
\[ W = F \int_{0}^{1} dx \]

5. **Substituting for Displacement**:
We need to express \( dx \) in terms of \( dt \):
\[ dx = v \, dt = 4t \, dt \]
Therefore, the work done becomes:
\[ W = 4 \int_{0}^{1} (4t) \, dt = 16 \int_{0}^{1} t \, dt \]

6. **Evaluate the Integral**:
The integral of \( t \) from 0 to 1 is:
\[ \int_{0}^{1} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]
Thus, substituting back:
\[ W = 16 \times \frac{1}{2} = 8 \, \text{J} \]

### Final Answer:
The work done in the first one second by the external force is:
\[ \boxed{8 \, \text{J}} \]