Question

An airplane is cruising along in a horizontal level flight at a constant velocity, heading due west. (a) If the weight of the plane is $2.6 \times 10^{4} \mathrm{N},$ what is the net force on the plane? (b) With what force does the air push upward on the plane?

Short answer

Answer: (a) The net force on the airplane during horizontal level flight with a constant velocity is 0 N. (b) The force of air pushing upward on the airplane (lift) is equal to the weight of the airplane, which is 2.6 × 10^4 N.

Step-by-step solution

Analyze the given information

We are given that the airplane is in horizontal level flight and moving with a constant velocity. We are also provided with the weight of the airplane, which is \(2.6 \times 10^4\,\mathrm{N}\).

Determine the net force on the airplane

Since the airplane is in horizontal level flight at a constant velocity, it means that there is no acceleration. According to Newton's first law of motion, an object in motion will stay in motion at a constant velocity unless acted upon by a net force. In this case, there is no net force acting on the airplane. Therefore, the net force is zero, even though there are forces acting on the plane. Net force = \(0\,\mathrm{N}\)

Determine the force of air pushing upward (lift)

In a horizontal level flight, the force of air pushing upward on the airplane, also known as lift, is equal in magnitude to the weight of the airplane but acts in the opposite direction (upward) to balance the downward gravitational force. This ensures that the net vertical force is zero, which is crucial for maintaining a level flight. Since we know the weight of the airplane, we can determine the lift force as follows: Lift force = \(2.6 \times 10^4\,\mathrm{N}\) (upward)

Present the results

(a) The net force on the airplane in horizontal level flight is \(0\,\mathrm{N}\). (b) The force of air pushing upward on the airplane (lift) is equal to the weight of the airplane, which is \(2.6 \times 10^4\,\mathrm{N}\).