Engine thrust 38 kN. What work is done by the engine in 20 seconds of flight at a speed of 1800 km / h?

Engine thrust 38 kN. What work is done by the engine in 20 seconds of flight at a speed of 1800 km / h? Calculate the engine power.

These tasks: F (traction force of the engine used) = 38 kN (38 * 10 ^ 3 N); t (duration of the engine’s work) = 20 s; V (flight speed) = 1800 km / h (500 m / s).

1) Power of the motor used: N = F * V = 38 * 10 ^ 3 * 500 = 19 * 10 ^ 6 W (19 MW).

2) Engine operation in 20 s: A = F * S = F * V * t = N * t = 19 * 10 ^ 6 * 20 = 380 * 10 ^ 6 J (380 MJ).

Answer: In 20 seconds of flight, the 19 MW engine performed work equal to 380 MJ.