What is the kinetic and potential energy of a Boeing 747 airliner weighing 2.48 106 N, flying at 268 m/s (i.e., 600 mi/h) at an altitude of 6.2 km (i.e., 20 103 ft)?
KE= 1/2 m v^2
GPe= mgh
To calculate the kinetic and potential energy of the Boeing 747 airliner, we need to use the formulas for kinetic energy and potential energy.
1. Kinetic Energy (KE):
The kinetic energy of an object is given by the formula:
KE = (1/2)mv^2
where m is the mass of the object and v is its velocity.
First, we need to find the mass of the airliner. We can use the formula:
mass = weight / acceleration due to gravity
Given weight = 2.48 * 10^6 N and acceleration due to gravity ≈ 9.8 m/s^2, we can calculate the mass as follows:
mass = 2.48 * 10^6 N / 9.8 m/s^2
After calculating the mass, we can use the given velocity (268 m/s) to find the kinetic energy using the formula above.
2. Potential Energy (PE):
The potential energy of an object at a certain height is given by the formula:
PE = mgh
where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.
First, we need to convert the height of the airliner from feet to meters. Given 1 ft ≈ 0.305 m, we can convert the height from 20 * 10^3 ft to meters by multiplying it by 0.305.
After converting the height to meters, we can use the calculated mass and the height to find the potential energy using the formula above.
Once we have both the kinetic energy and potential energy, we can provide the requested values.
Please note that the above calculations assume no other external factors affecting the aircraft, such as air resistance or changes in energy due to engines.