A force acts on a 3.0 gm particle in such a way that the position of the particle as a function of time is given by ` x=3t-4t^(2)+t^(3)`, where xx is in metres and t is in seconds. The work done during the first 4 seconds is

Options

A. 5.28 J
B. 450 mJ
C. 490 mJ
D. 530 mJ

Answer

To solve the problem, we need to find the work done on a particle whose position as a function of time is given by \( x(t) = 3t - 4t^2 + t^3 \). The mass of the particle is 3 grams, which we will convert to kilograms for our calculations.

### Step-by-Step Solution:

1. **Convert mass from grams to kilograms**:
\[ m = 3 \text{ gm} = 0.003 \text{ kg} \]

2. **Find the velocity**:
The velocity \( v(t) \) is the first derivative of the position function \( x(t) \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t - 4t^2 + t^3) = 3 - 8t + 3t^2 \]

3. **Find the acceleration**:
The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(3 - 8t + 3t^2) = -8 + 6t \]

4. **Calculate the force**:
Using Newton's second law, \( F = ma \):
\[ F(t) = m \cdot a(t) = 0.003 \cdot (-8 + 6t) \]

5. **Calculate the work done**:
The work done \( W \) is given by the integral of the force over the displacement. Since \( ds = dx \), we have:
\[ W = \int_0^4 F(t) \cdot v(t) \, dt \]
Substituting \( F(t) \) and \( v(t) \):
\[ W = \int_0^4 (0.003)(-8 + 6t)(3 - 8t + 3t^2) \, dt \]

6. **Expand the integrand**:
First, we expand \( (-8 + 6t)(3 - 8t + 3t^2) \):
\[ = -24 + 64t - 24t^2 + 18t - 48t^2 + 18t^3 \]
Combining like terms:
\[ = -24 + 82t - 72t^2 + 18t^3 \]

7. **Integrate the expression**:
Now we integrate:
\[ W = 0.003 \int_0^4 (-24 + 82t - 72t^2 + 18t^3) \, dt \]
The integral can be computed as follows:
\[ = 0.003 \left[ -24t + 41t^2 - 24t^3 + \frac{18}{4}t^4 \right]_0^4 \]
Evaluating from 0 to 4:
\[ = 0.003 \left[ -24(4) + 41(16) - 24(64) + 4.5(256) \right] \]
\[ = 0.003 \left[ -96 + 656 - 1536 + 1152 \right] \]
\[ = 0.003 \left[ -96 + 656 - 1536 + 1152 \right] = 0.003 \left[ 176 \right] \]
\[ = 0.528 \text{ Joules} \]

8. **Convert to milliJoules**:
\[ W = 528 \text{ mJ} \]

### Final Answer:
The work done during the first 4 seconds is \( 528 \text{ mJ} \).