An airplane of mass 2800 Kg has just lifted off the runway. It is gaining altitude at a constant 2.3 M/s while the horizontal component of its velocity is increasing at a rate of 0.86 M/s2. Assume g= 9.81 M/s2. (a) Find the direction of the force exerted on the airplane by the air. (b) Find the horizontal and vertical components of the planes acceleration if the force due to the air has the same magnitude but has a direction of 2.0ᴼ closer to the vertical than its description in part (a).
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To find the direction of the force exerted on the airplane by the air, we need to consider the forces acting on the airplane. In this case, the airplane is gaining altitude, which means there must be a force pushing it upwards against gravity.
(a) The force exerted on the airplane by the air can be broken down into two components: the vertical force and the horizontal force. The vertical force is responsible for the airplane's upwards acceleration and can be found using Newton's second law, F = ma, where F is the force, m is the mass of the airplane, and a is the upward acceleration.
From the given information, we know that the upward acceleration is 2.3 m/s and the mass of the airplane is 2800 kg. Plugging these values into the equation, we have:
F = m * a
F = 2800 kg * 2.3 m/s^2
F = 6440 N (upwards)
Therefore, the force exerted on the airplane by the air is 6440 N in the upward direction.
(b) To find the horizontal and vertical components of the airplane's acceleration, we need to consider the direction of the force due to the air. According to the question, the force due to the air has the same magnitude as in part (a) but is 2.0ᴼ closer to the vertical.
Let's break down the force due to the air into horizontal and vertical components. We'll label the horizontal component as F_horizontal and the vertical component as F_vertical.
Using trigonometry, we can find the values of F_horizontal and F_vertical:
F_horizontal = F * sin(2.0ᴼ)
F_vertical = F * cos(2.0ᴼ)
Substituting the value of F (6440 N), we have:
F_horizontal = 6440 N * sin(2.0ᴼ)
F_vertical = 6440 N * cos(2.0ᴼ)
Calculating these values, we get:
F_horizontal ≈ 222 N (in the direction closer to the vertical)
F_vertical ≈ 6432 N (in the upward direction)
Therefore, the horizontal component of the airplane's acceleration is approximately 222 N, directed closer to the vertical, and the vertical component of the acceleration is approximately 6432 N, directed upwards.