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The question asks for the power rating of an electric motor, given that it does **45.0 kJ** of work in **30.0 s**. 

The formula for power is:

\[
\text{Power} = \frac{\text{Work}}{\text{Time}}
\]

Where:
- Work = 45.0 kJ = \( 45.0 \times 10^3 \) J (since 1 kJ = 1000 J),
- Time = 30.0 s.

Now, let's calculate the power in watts and then convert it to kilowatts (1 kW = 1000 W).

\[
\text{Power} = \frac{45.0 \times 10^3 \, \text{J}}{30.0 \, \text{s}} = 1500 \, \text{W}
\]

Converting to kilowatts:

\[
\text{Power} = \frac{1500 \, \text{W}}{1000} = 1.5 \, \text{kW}
\]

So, the power rating of the motor is **1.5 kW**.

Would you like more details on this calculation or have any questions?

Here are some related questions:
1. What is the unit of power in the International System of Units (SI)?
2. How do you convert joules to kilojoules?
3. How does the duration of time affect the power rating of a device?
4. What is the relation between work, energy, and power?
5. Can power ever be negative, and what would that mean in this context?

**Tip:** When converting between units, always make sure the conversion factors (like 1 kJ = 1000 J) are applied correctly to avoid errors.

An electric motor is capable of doing 45.0 kJ of work in 30.0 s. What is the power rating of the motor in kilowatts?

This problem demonstrates how to calculate the power rating of an electric motor. Given 45.0 kJ of work done in 30.0 seconds, the problem involves using the power formula to convert work into kilowatts. Suitable for high school physics or engineering students.

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