The position of a Particle of Mass 4 g,acted upon by a constant force is given by x=`4t^(2)`+t,where x is in metre and t in second .The work done during the first 2 seconds is

Options

A. 128 mJ
B. 512 mJ
C. 576 mJ
D. 144mJ

Explanation

To solve the problem of finding the work done on a particle of mass 4 g, given its position function \( x = 4t^2 + t \), we can follow these steps:

### Step 1: Differentiate the position function to find velocity
The position of the particle is given by:
\[ x(t) = 4t^2 + t \]
To find the velocity \( v(t) \), we differentiate \( x(t) \) with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(4t^2 + t) = 8t + 1 \]

### Step 2: Differentiate the velocity function to find acceleration
Next, we differentiate the velocity function to find the acceleration \( a(t) \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(8t + 1) = 8 \]
Since the acceleration is constant, we can conclude that \( a = 8 \, \text{m/s}^2 \).

### Step 3: Calculate the force acting on the particle
Using Newton's second law, the force \( F \) acting on the particle can be calculated as:
\[ F = m \cdot a \]
The mass \( m \) is given as 4 g, which we need to convert to kilograms:
\[ m = 4 \, \text{g} = 0.004 \, \text{kg} \]
Now, substituting the values:
\[ F = 0.004 \, \text{kg} \cdot 8 \, \text{m/s}^2 = 0.032 \, \text{N} \]

### Step 4: Calculate the displacement during the first 2 seconds
To find the displacement \( \Delta x \) during the first 2 seconds, we evaluate the position function at \( t = 2 \) seconds:
\[ x(2) = 4(2^2) + 2 = 4(4) + 2 = 16 + 2 = 18 \, \text{m} \]
At \( t = 0 \):
\[ x(0) = 4(0^2) + 0 = 0 \, \text{m} \]
Thus, the total displacement during the first 2 seconds is:
\[ \Delta x = x(2) - x(0) = 18 - 0 = 18 \, \text{m} \]

### Step 5: Calculate the work done
The work done \( W \) by the force over the displacement is given by:
\[ W = F \cdot \Delta x \]
Substituting the values we found:
\[ W = 0.032 \, \text{N} \cdot 18 \, \text{m} = 0.576 \, \text{J} \]

### Final Answer
The work done during the first 2 seconds is:
\[ \boxed{0.576 \, \text{J}} \]