Moment of inertia of a body about a given axis is `1.5 kg m^(2)`. Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration of `20 rad//s^(2)` must be applied about the axis for a duration of _________ (in sec).

Explanation

To solve the problem step-by-step, we will follow these steps:

### Step 1: Identify the given values
- Moment of inertia (I) = 1.5 kg m²
- Initial angular velocity (ω_initial) = 0 rad/s (since the body is at rest)
- Rotational kinetic energy (KE) = 1200 J
- Angular acceleration (α) = 20 rad/s²

### Step 2: Use the formula for rotational kinetic energy
The formula for rotational kinetic energy is given by:
\[ KE = \frac{1}{2} I \omega^2 \]
We need to find the final angular velocity (ω_final) using the given kinetic energy and moment of inertia.

### Step 3: Rearrange the formula to solve for ω_final
Substituting the known values into the kinetic energy formula:
\[ 1200 = \frac{1}{2} \times 1.5 \times \omega^2 \]
This simplifies to:
\[ 1200 = 0.75 \omega^2 \]

### Step 4: Solve for ω_final
Now, we can solve for ω:
\[ \omega^2 = \frac{1200}{0.75} = 1600 \]
Taking the square root gives:
\[ \omega = \sqrt{1600} = 40 \text{ rad/s} \]

### Step 5: Use the angular motion equation to find time
We will use the angular motion equation:
\[ \omega_{final} = \omega_{initial} + \alpha t \]
Substituting the known values:
\[ 40 = 0 + 20t \]

### Step 6: Solve for time (t)
Rearranging the equation gives:
\[ t = \frac{40}{20} = 2 \text{ seconds} \]

### Final Answer
The duration for which the angular acceleration must be applied is **2 seconds**.

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