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A child leaves her book bag on a slide. the bag, which is at the top of the slide, starts from rest and reaches the bottom in 1.47 s. the mass of the book bag is 2.45 kg, the length of the slide is 2.50 m and the angle of incline is 30.5°. (assume the +x-axis to be parallel to and down the slide. for all values, enter the magnitude only.)

Respuesta :

meerkat18
From kinematic relation d = vi t + 1/2 at^2 , a = 2d/t^2 since it starts from rest . Acceleration does not depend on the mass. a = 2* 2.5 / (1.47)^2 = 2.3139 m/s^2.
Acceleration down the slide is due to the parallel component of gravity force. Friction opposes this force and acts up the inclined plane. mg sin theta - f = ma. Friction = f = (2.45 * 9.8) sin 30.5 - (2.45)(2.3139) = 6.5169 N. 
Normal force = mg cos 30.5 = (2.45*9.8)* cos 30.5 = 20.6877 N. Friction = uk N. coefficient of friction =   = friction / N = 6.5169 / 20.6877 = 0.32. 
When it reaches the bottom of the slide, vf = vi + at = 0 + ( 2.3139)(1.47) = 3.4014 m/s = velocity at the bottom

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