The displacement of a body of mass `2 kg` varies with time `t` as `S = t^(2) + 2t`, where `S` is in seconds. The work done by all the forces acting on the body during the time interval `t = 2s` to `t = 4s` is

Options

A. 36 J
B. 64 J
C. 100 J
D. 120 J

Answer

To solve the problem, we need to find the work done by all the forces acting on a body of mass 2 kg during the time interval from t = 2 seconds to t = 4 seconds. The displacement of the body is given by the equation \( S = t^2 + 2t \).

### Step-by-Step Solution:

1. **Identify the given information:**
- Mass of the body, \( m = 2 \, \text{kg} \)
- Displacement equation, \( S(t) = t^2 + 2t \)
- Time interval: \( t = 2 \, \text{s} \) to \( t = 4 \, \text{s} \)

2. **Calculate the velocity:**
- Velocity \( v \) is the rate of change of displacement with respect to time, given by:
\[ v = \frac{dS}{dt} \]
- Differentiate \( S(t) \):
\[ v = \frac{d}{dt}(t^2 + 2t) = 2t + 2 \]

3. **Find the final velocity at \( t = 4 \, \text{s} \):**
- Substitute \( t = 4 \):
\[ v_f = 2(4) + 2 = 8 + 2 = 10 \, \text{m/s} \]

4. **Find the initial velocity at \( t = 2 \, \text{s} \):**
- Substitute \( t = 2 \):
\[ v_i = 2(2) + 2 = 4 + 2 = 6 \, \text{m/s} \]

5. **Calculate the change in kinetic energy:**
- Kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2} mv^2 \]
- Change in kinetic energy \( \Delta KE \) is:
\[ \Delta KE = KE_f - KE_i = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \]
- Substitute \( m = 2 \, \text{kg} \), \( v_f = 10 \, \text{m/s} \), and \( v_i = 6 \, \text{m/s} \):
\[ \Delta KE = \frac{1}{2} \cdot 2 \cdot (10^2) - \frac{1}{2} \cdot 2 \cdot (6^2) \]
\[ = 1 \cdot 100 - 1 \cdot 36 = 100 - 36 = 64 \, \text{J} \]

6. **Conclusion:**
- The work done by all the forces acting on the body during the time interval from \( t = 2 \, \text{s} \) to \( t = 4 \, \text{s} \) is \( 64 \, \text{J} \).

### Final Answer:
The work done by all the forces is **64 Joules**.