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`

Explanation

To find the work done by all the forces acting on the body during the time interval from \( t = 2s \) to \( t = 4s \), we can follow these steps:

### Step 1: Find the Displacement Function
The displacement \( S \) is given by:
\[ S = t^2 + 2t \]

### Step 2: Calculate the Velocity
To find the velocity \( v \), we differentiate the displacement \( S \) with respect to time \( t \):
\[ v = \frac{dS}{dt} = \frac{d}{dt}(t^2 + 2t) = 2t + 2 \]

### Step 3: Calculate the Acceleration
Next, we find the acceleration \( a \) by differentiating the velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(2t + 2) = 2 \]

### Step 4: Calculate Work Done
The work done \( W \) by the forces acting on the body can be calculated using the formula:
\[ W = \int_{t_1}^{t_2} m \cdot a \cdot v \, dt \]
where \( m \) is the mass of the body, \( a \) is the acceleration, and \( v \) is the velocity.

Given:
- Mass \( m = 2 \, \text{kg} \)
- Acceleration \( a = 2 \) (constant)
- Velocity \( v = 2t + 2 \)

Now we can substitute these values into the work done formula:
\[ W = \int_{2}^{4} 2 \cdot (2) \cdot (2t + 2) \, dt \]
\[ W = \int_{2}^{4} 4(2t + 2) \, dt \]
\[ W = \int_{2}^{4} (8t + 8) \, dt \]

### Step 5: Evaluate the Integral
Now we evaluate the integral:
\[ W = \left[ 4t^2 + 8t \right]_{2}^{4} \]
Calculating at the limits:
\[ W = \left( 4(4^2) + 8(4) \right) - \left( 4(2^2) + 8(2) \right) \]
\[ W = \left( 4(16) + 32 \right) - \left( 4(4) + 16 \right) \]
\[ W = (64 + 32) - (16 + 16) \]
\[ W = 96 - 32 = 64 \, \text{J} \]

### Final Answer
The work done by all the forces acting on the body during the time interval from \( t = 2s \) to \( t = 4s \) is \( 64 \, \text{J} \).

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