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 find the work done by all the forces acting on a particle of mass 3 kg moving along the x-axis, we can follow these steps:

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

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

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

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

### Step 5: Calculate the work done over the time interval
The work done \( W \) by the force over a displacement can be calculated using the formula:
\[ W = F \cdot d \]
where \( d \) is the total displacement over the time interval from \( t = 0 \) to \( t = 3 \) seconds.

### Step 6: Find the displacement over the time interval
To find the displacement, we evaluate the position function at \( t = 3 \) and \( t = 0 \):
- At \( t = 3 \):
\[ x(3) = 2(3^2) + 5 = 2(9) + 5 = 18 + 5 = 23 \text{ m} \]
- At \( t = 0 \):
\[ x(0) = 2(0^2) + 5 = 5 \text{ m} \]

Thus, the total displacement \( d \) is:
\[ d = x(3) - x(0) = 23 \text{ m} - 5 \text{ m} = 18 \text{ m} \]

### Step 7: Calculate the work done
Now we can calculate the work done using the force and the displacement:
\[ W = F \cdot d = 12 \text{ N} \cdot 18 \text{ m} = 216 \text{ J} \]

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