# Power Output of Engine from Torque and Rotational Speed Given: - Torque, \( \tau = 180 \) Nm - Rotational speed, \( N = 250 \) rpm (reutions per minute) We'll find the power output in watts (W). --- ## Step 1: Convert Rotational Speed to Radians per Second The angular velocity \( \omega \) in radians per second is calculated as: \[ \omega = 2\pi N / 60 \] where \( N \) is in rpm. \[ \omega = 2\pi \times 250 / 60 \] \[ \omega = 2\pi \times 41.6667 \] \[ \omega \approx 2 \times 3.1416 \times 41.6667 \] \[ \omega \approx 6.2832 \times 41.6667 \] \[ \omega \approx 261.80~\text{rad/s} \] --- ## Step 2: Calculate Power Output Power (\( P \)) is given by: \[ P = \tau \times \omega \] Substitute values: \[ P = 180~\text{Nm} \times 261.80~\text{rad/s} \] \[ P = 47,124~\text{W} \] --- ## Step 3: Express Power in Kilowatts \[ P = 47,124~\text{W} = 47.1~\text{kW} \] --- ## **Final Answer** **The power output of the engine is \( \boxed{47,124~\text{W}} \) or \( \boxed{47.1~\text{kW}} \).**