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. 530 mJ
B. 490 mJ
C. 450 mJ
D. 2.28J

Explanation

To solve the problem of finding the work done on a 3.0 gm particle whose position as a function of time is given by \( x(t) = 3t - 4t^2 + t^3 \), we will follow these steps:

### Step 1: Differentiate the position function to find the velocity function.
The velocity \( v(t) \) is the derivative of the position function \( x(t) \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t - 4t^2 + t^3) \]
Calculating the derivative:
\[ v(t) = 3 - 8t + 3t^2 \]

### Step 2: Calculate the initial velocity \( v(0) \).
Substituting \( t = 0 \) into the velocity function:
\[ v(0) = 3 - 8(0) + 3(0)^2 = 3 \, \text{m/s} \]

### Step 3: Calculate the velocity at \( t = 4 \) seconds, \( v(4) \).
Substituting \( t = 4 \) into the velocity function:
\[ v(4) = 3 - 8(4) + 3(4)^2 \]
Calculating:
\[ v(4) = 3 - 32 + 48 = 19 \, \text{m/s} \]

### Step 4: Calculate the mass of the particle in kg.
The mass is given as 3.0 grams, which we convert to kilograms:
\[ m = 3.0 \, \text{g} = 3.0 \times 10^{-3} \, \text{kg} \]

### Step 5: Use the work-energy theorem to calculate the work done.
The work done \( W \) is given by the change in kinetic energy:
\[ W = \frac{1}{2} m (v_f^2 - v_i^2) \]
Where \( v_f = v(4) \) and \( v_i = v(0) \):
\[ W = \frac{1}{2} (3.0 \times 10^{-3}) \left(19^2 - 3^2\right) \]

### Step 6: Calculate \( v_f^2 \) and \( v_i^2 \).
Calculating \( v_f^2 \) and \( v_i^2 \):
\[ v_f^2 = 19^2 = 361 \]
\[ v_i^2 = 3^2 = 9 \]

### Step 7: Substitute back into the work equation.
Now substituting back:
\[ W = \frac{1}{2} (3.0 \times 10^{-3}) (361 - 9) \]
Calculating the difference:
\[ W = \frac{1}{2} (3.0 \times 10^{-3}) (352) \]

### Step 8: Final calculation of work done.
Calculating the work done:
\[ W = \frac{1}{2} \times 3.0 \times 10^{-3} \times 352 = 1.5 \times 352 \times 10^{-3} \]
Calculating:
\[ W = 528 \times 10^{-3} \, \text{J} = 0.528 \, \text{J} = 528 \, \text{mJ} \]

### Final Answer:
The work done during the first 4 seconds is approximately **528 mJ**.

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