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Unformatted text preview: W = mg = (15.0kg) (9.80m/s 2 ) = 147N , F x = Wsin θ – f k = m a x = 0 , so f k = (147N) sin14.0 o = 35.6N , and F y = n – Wcos θ = m a y = 0 , so n = (147N) cos14.0 o = 143N . W = 147N , f k = 35.6N , n = 143N . θ n f W y x 3. Consider a mass, m = 20.0kg, sliding down an incline at an angle of θ = 17.0 o , as shown above. There is a force of kinetic friction of 30.0N acting on the mass. Draw a free-body diagram for the mass. Determine numerical values for all of the forces acting on the mass and determine the acceleration of the mass. With the same vectors above we have W = mg = (20.0kg) (9.80m/s 2 ) = 196N , F x = Wsin θ – f k = (196N) sin17.0 o – 30.0N = 27.3N = ma x , so a x = (27.3N) / (20.0kg) = 1.37m/s 2 , and F y = n – Wcos θ = m a y = 0 , so n = (196N) cos17.0 o = 187N . W = 196N , f k = 30.0N , n = 187N, a x = 1.37m/s 2 ....
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- Spring '11