The displacement of an object of mass 3 kg is given by the relation `S=(1)/(3)t^(2)`, where t is time in seconds. If the work done by the net force on the object in 2 s is `(p)/(q)` joule, where p and q are smallest integer values, then what is the value of `p+q`?

Answer

To solve the problem, we need to determine the work done by the net force on the object over a time period of 2 seconds, given the displacement function.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Mass of the object, \( m = 3 \, \text{kg} \)
- Displacement function, \( S(t) = \frac{1}{3} t^2 \)

2. **Calculate the Velocity:**
- Velocity is the first derivative of displacement with respect to time:
\[ v(t) = \frac{dS}{dt} = \frac{d}{dt}\left(\frac{1}{3} t^2\right) = \frac{2}{3} t \]

3. **Calculate the Acceleration:**
- Acceleration is the derivative of velocity with respect to time:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{2}{3} t\right) = \frac{2}{3} \]
- The acceleration is constant at \( \frac{2}{3} \, \text{m/s}^2 \).

4. **Calculate the Net Force:**
- Using Newton's second law, \( F = m \cdot a \):
\[ F_{\text{net}} = 3 \, \text{kg} \cdot \frac{2}{3} \, \text{m/s}^2 = 2 \, \text{N} \]

5. **Calculate the Displacement in 2 Seconds:**
- Substitute \( t = 2 \) seconds into the displacement function:
\[ S(2) = \frac{1}{3} (2^2) = \frac{1}{3} \cdot 4 = \frac{4}{3} \, \text{m} \]

6. **Calculate the Work Done:**
- Work done is given by the formula \( W = F \cdot D \):
\[ W = 2 \, \text{N} \cdot \frac{4}{3} \, \text{m} = \frac{8}{3} \, \text{J} \]

7. **Express Work Done in the Form \( \frac{p}{q} \):**
- Here, \( p = 8 \) and \( q = 3 \).

8. **Calculate \( p + q \):**
- Therefore, \( p + q = 8 + 3 = 11 \).

### Final Answer:
The value of \( p + q \) is \( 11 \).

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