A foce acts on a 30.g 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 x is in metre and t in second. The work done during the first 4s is

Options

A. 570mJ
B. 450mJ
C. 490mJ
D. 528mJ

Answer

To find the work done on a 30 g particle during the first 4 seconds, we can follow these steps:

### Step 1: Determine the position function
The position of the particle is given by the equation:
\[ x(t) = 3t - 4t^2 + t^3 \]

### Step 2: Calculate the velocity function
The velocity \( v(t) \) is the derivative of the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t - 4t^2 + t^3) \]
Calculating the derivative:
\[ v(t) = 3 - 8t + 3t^2 \]

### Step 3: Find the initial and final velocities
Now, we need to find the velocities at \( t = 0 \) seconds and \( t = 4 \) seconds.

- For \( t = 0 \):
\[ v(0) = 3 - 8(0) + 3(0)^2 = 3 \, \text{m/s} \]

- For \( t = 4 \):
\[ v(4) = 3 - 8(4) + 3(4)^2 = 3 - 32 + 48 = 19 \, \text{m/s} \]

### Step 4: Convert mass to kilograms
The mass of the particle is given as 30 g. We need to convert this to kilograms:
\[ m = 30 \, \text{g} = 30 \times 10^{-3} \, \text{kg} = 0.030 \, \text{kg} \]

### Step 5: Calculate the change in kinetic energy
The work done on the particle is equal to the change in kinetic energy (\( \Delta KE \)):
\[ \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \]
Substituting the values we found:
\[ \Delta KE = \frac{1}{2} (0.030) (19^2) - \frac{1}{2} (0.030) (3^2) \]

Calculating each term:
\[ \Delta KE = \frac{1}{2} (0.030) (361) - \frac{1}{2} (0.030) (9) \]
\[ = 0.015 \times 361 - 0.015 \times 9 \]
\[ = 5.415 - 0.135 = 5.280 \, \text{J} \]

### Step 6: Convert to milliJoules
To express the work done in milliJoules:
\[ 5.280 \, \text{J} = 5280 \, \text{mJ} \]

### Final Answer
The work done during the first 4 seconds is:
\[ \text{Work Done} = 5280 \, \text{mJ} \]

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