A particle of mass 3 kg is moving along x - axis and its position at time t is given by equation `x=(2t^(2)+5)m`. Work done by all the force acting on it in time interval `t=0` to `t=3s` is

Options

A. 144 J
B. 72 J
C. 108 J
D. 216 J

Answer

To solve the problem step by step, we will follow the outlined process to find the work done by all forces acting on the particle during the time interval from \( t = 0 \) to \( t = 3 \) seconds.

### Step 1: Identify the position function
The position of the particle is given by the equation:
\[ x(t) = 2t^2 + 5 \]

### Step 2: Calculate the velocity
To find the velocity, we differentiate the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(2t^2 + 5) = 4t \]

### Step 3: Calculate the acceleration
Next, we differentiate the velocity function to find the acceleration:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(4t) = 4 \, \text{m/s}^2 \]
The acceleration is constant at \( 4 \, \text{m/s}^2 \).

### Step 4: Calculate the force acting on the particle
Using Newton's second law, \( F = m \cdot a \), where \( m = 3 \, \text{kg} \):
\[ F = 3 \, \text{kg} \cdot 4 \, \text{m/s}^2 = 12 \, \text{N} \]

### Step 5: Calculate the displacement during the time interval
Now we need to find the displacement from \( t = 0 \) to \( t = 3 \) seconds. We calculate the initial and final positions:
- At \( t = 0 \):
\[ x(0) = 2(0)^2 + 5 = 5 \, \text{m} \]
- At \( t = 3 \):
\[ x(3) = 2(3)^2 + 5 = 2 \cdot 9 + 5 = 18 + 5 = 23 \, \text{m} \]

The displacement \( \Delta x \) is:
\[ \Delta x = x(3) - x(0) = 23 \, \text{m} - 5 \, \text{m} = 18 \, \text{m} \]

### Step 6: Calculate the work done
The work done \( W \) by the force is given by:
\[ W = F \cdot \Delta x \]
Substituting the values:
\[ W = 12 \, \text{N} \cdot 18 \, \text{m} = 216 \, \text{J} \]

### Final Answer
The work done by all the forces acting on the particle during the time interval from \( t = 0 \) to \( t = 3 \) seconds is:
\[ \boxed{216 \, \text{J}} \]