NASA Guide to Engines

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NASA Guide to Engines ( nasa-guide-engines )

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x-direction and two units in the positive y-direction (illustrated below at the left). There is another way to obtain the component parts of a given vector. If, for example, the magnitude of the acceleration and the angle of motion are known, then the component parts of the acceleration can be determined by using trigonometry (illustrated beloFw=a(t3ith+e2rji)gNht). y |F| Fx = 3 N F = (3i + 2j) N Fy = 2 N y ax = a cosθ y ax = a cosθ a =asinθ |a| ay =asinθ x θ Newton’s Second Law of Motion x y |F| Fx = 3 N Fy = 2 N x F= 32+22= 13N F= 32+22= 13N y Newton’s second law answers the question of what happens to an object that has a nonzero net force axcting on it. The following explanations are for a constant mass. One observation from Newton’s second law is that the acceleration of an object is directly proportional to the net force acting on it. For example, if you are pushing a block of metal across a frictionless horizontal surface with some horizontal force F, the block moves with some acceleration a. If you apply a force of 3F on the same metal block, the acceleration of the block also triples. Another observation is that the acceleration of an object is inversely proportional to its mass. Using the block example from above, if one doubles the mass of the metal block and applies the same force F, the acceleration of the block will be a . 2 Newton’s second law of motion can also be observed in the motion of an aircraft. If an airplane is not traveling at a constant velocity  Dv ≠ 0  , then the airplane is accelerating or decelerating. For example, before takeoff, Dt  the airplane is stationary (Newton’s first law). The throttle is then increased to overcome the inertia of the plane. Because of the positive net force (thrust), the airplane accelerates (Newton’s second law). Example Problem: A jet with a mass of 250,000. kg is cruising at 241.30 m/s at a constant altitude. If the afterburners apply an additional thrust of 50,000. N, what speed does the jet reach in 100. s? Solution: This problem can be solved using Newton’s second law of motion, F = m Dv . F = m vf − vi , where F is the Dt tf −ti additional thrust (note that until the afterburners are turned on, the net force equals zero), vf is the velocity when tf = 100. s, and vi = 241.30 m/h when ti = 0 s. The final velocity can be determined: 50,000. N = (250,000. kg) (vf − 241.30) m/s (100.−0) s vf = 261 m/s |a| θ 50

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