A 17,000-kg airplane lands with a speed of 82 m/s on a stationary aircraft carrier deck that is 115 m long. find the work done by nonconservative forces in stopping the plane

Answers

Answer 1

Answer:

Final answer:

The work done by nonconservative forces in stopping the 17,000-kilogram airplane landing at a speed of 82 m/s is 57,062,000 Joules. This is calculated by the change in kinetic energy of the airplane when it lands and comes to a stop.

Explanation:

The question refers to the concept of work-energy theorem in Physics, especially involving non-conservative forces. The airplane is initially moving and finally comes to rest. Its initial kinetic energy (KE) gets transferred to work done by nonconservative forces, which in this scenario includes friction due to the aircraft carrier deck and air resistance.

The initial kinetic energy of the plane is calculated using the formula 1/2 * m * v^2 where 'm' is the mass of the plane and 'v' is its speed. So, the initial kinetic energy of the plane is 1/2 * 17,000 kg * (82 m/s)^2 = 57,062,000 Joules. When the plane comes to rest, its final kinetic energy is 0. As per the work-energy theorem, the work done by nonconservative forces is equal to the change in the kinetic energy. Therefore, the work done by nonconservative forces in stopping the plane = Initial KE - Final KE = 57,062,000 Joules - 0 = 57,062,000 Joules.

Learn more about Work-energy theorem here:

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Answers

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Explanation:

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