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In an electric motor, energy conversion is the process where electrical energy is transformed into mechanical energy. This is the primary function of an electric motor and it is crucial for the motor's operation. In the given problem, the motor consumes a total of 9,000 Joules of electrical energy. However, this entire amount is not converted into useful work. Instead, only two-thirds of it is utilized effectively while the remaining one-third is unfortunately lost to heat and other internal processes. This means the useful energy available for the motor's operation is calculated as \( \frac{2}{3} \times 9,000 = 6,000 \text{ J} \). This highlights the importance of energy efficiency in electric motors, emphasizing the need to minimize energy losses to improve performance.
Understanding how energy conversion happens helps optimize the systems using electric motors, making them more efficient and effective for various applications.

Torque in an electric motor is a measure of the rotational force that the motor generates. This is a crucial element as it determines how much "push" the motor can provide to do its work. Torque (\( \tau \)) can be calculated using the relationship between power output, torque, and angular velocity, which is expressed as \( P = \tau \omega \), where \( P \) is power and \( \omega \) is angular velocity.
To find torque, we rearrange the formula to obtain \( \tau = \frac{P}{\omega} \). In the example provided, the power output was 100 watts, and the angular velocity was approximately 261.8 radians per second. Plugging these values into the formula produces a torque of about 0.382 Newton-meters (Nm). This calculation is vital for understanding the mechanical capability of the motor to perform a rotating task.

Angular velocity is how fast an object rotates or revolves relative to another point, often the center of a circle. In motors, it signifies how quickly the motor shaft spins. It is imperative to convert the motor's speed from revolutions per minute (rpm) to radians per second to use it in torque calculations.
To perform this conversion, we use the formula \( \omega = \frac{2\pi n}{60} \), where \( n \) is the speed in rpm. For an electric motor running at 2500 rpm, the angular velocity is calculated as \( \omega = \frac{2\pi \times 2500}{60} \approx 261.8 \text{ rad/s} \). Understanding this concept helps when you need to relate rotational speed to other properties like torque and power, providing a comprehensive look into the motor's functioning.

Power output is a crucial characteristic of an electric motor, indicating how much work the motor can perform in a certain time frame. It is the useful energy output per unit of time, often measured in watts (W). In this context, power output is derived from the useful energy that has not been lost as heat or internal energy.
In the exercise, the useful available energy is 6,000 Joules in 60 seconds, leading to a power output of \( P = \frac{6000 \text{ J}}{60 \text{ s}} = 100 \text{ W} \). Calculating power output is essential for determining the efficiency and capacity of the motor to perform its intended function, such as moving a load or driving a mechanical system.