A 1.05 g samara (the winged fruit of a maple tree), falls towardthe ground with
ID: 1665589 • Letter: A
Question
A 1.05 g samara (the winged fruit of a maple tree), falls towardthe ground with a constant speed of 1.1 m/s. a) what is the force of the air resistance exerted on thesamara? b)if the constant speed of the descent os greater than 1.1m/s,is the force of the air resistance greater than, less than or thesame as in part A? Explain. a) what is the force of the air resistance exerted on thesamara? b)if the constant speed of the descent os greater than 1.1m/s,is the force of the air resistance greater than, less than or thesame as in part A? Explain.Explanation / Answer
a. Since the friut is fallingwith constant speed towards ground, no net force is acting upon it( else it would change its speed of direction). Fruit is actedupon by two forces, its weight working in downwards direction andair resistance in upwards direction. For net force to be zero Weight + airresistance = 0 m *g + Fair = 0 Resistance due toair Fair = -m * g - ve sign indicates that the air resistanceworks in upwards direction. Fair = -1.05 * 10-3 * 9.8 = - 1.029* 10-2 N b. As long as the speed is constant,the upwards force will equal to weight as calculated in partA. = - 1.029* 10-2 N b. As long as the speed is constant,the upwards force will equal to weight as calculated in partA.
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