Displacement of a particle of mass 2 kg varies with time as `s=(2t^(2)-2t + 10)m`. Find total work done on the particle in a time interval from `t=0` to `t=2s`.

Explanation

To find the total work done on the particle in the time interval from \( t = 0 \) to \( t = 2 \) seconds, we can follow these steps:

### Step 1: Write down the displacement equation
The displacement \( s \) of the particle is given by:
\[ s(t) = 2t^2 - 2t + 10 \]

### Step 2: Differentiate the displacement to find velocity
To find the velocity \( v \), we differentiate the displacement \( s \) with respect to time \( t \):
\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(2t^2 - 2t + 10) = 4t - 2 \]

### Step 3: Calculate initial and final velocities
Now, we will calculate the initial velocity at \( t = 0 \) and the final velocity at \( t = 2 \):
- At \( t = 0 \):
\[ v(0) = 4(0) - 2 = -2 \, \text{m/s} \]
- At \( t = 2 \):
\[ v(2) = 4(2) - 2 = 6 \, \text{m/s} \]

### Step 4: Use the work-energy theorem
The work done on the particle is equal to the change in kinetic energy:
\[ W = \Delta KE = KE_{final} - KE_{initial} \]
Where:
\[ KE = \frac{1}{2}mv^2 \]
Given that the mass \( m = 2 \, \text{kg} \):
- Initial kinetic energy \( KE_{initial} \):
\[ KE_{initial} = \frac{1}{2} \times 2 \times (-2)^2 = \frac{1}{2} \times 2 \times 4 = 4 \, \text{J} \]
- Final kinetic energy \( KE_{final} \):
\[ KE_{final} = \frac{1}{2} \times 2 \times (6)^2 = \frac{1}{2} \times 2 \times 36 = 36 \, \text{J} \]

### Step 5: Calculate the total work done
Now we can find the total work done:
\[ W = KE_{final} - KE_{initial} = 36 \, \text{J} - 4 \, \text{J} = 32 \, \text{J} \]

### Final Answer
The total work done on the particle in the time interval from \( t = 0 \) to \( t = 2 \) seconds is:
\[ \boxed{32 \, \text{J}} \]