# HW06Sol - W = mg =(15.0kg(9.80m/s 2 = 147N F x = Wsin θ...

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Jim Guinn’s PHYS1111 Assignment #6 Solutions Due Wed., Feb. 10, 2010 For all of these problems you might like to know that the weight of an object is equal to its mass times the acceleration due to gravity, i.e. W = m g . 1. Consider a mass, m = 10.0kg, at rest on an incline at an angle of θ = 15.0 o , as shown above. Draw a free-body diagram for the mass. Determine numerical values for all of the forces acting on the mass. See the vectors added above. For a body at rest we have W = mg = (10.0kg) (9.80m/s 2 ) = 98.0N , F x = Wsin θ – f s = m a x = 0 , so f s = (98.0N) sin15.0 o = 25.4N , and F y = n – Wcos θ = m a y = 0 , so n = (98.0N) cos15.0 o = 94.7N . W = 98.0N , f s = 25.4N , n = 94.7N . 2. Consider a mass, m = 15.0kg, sliding down with constant velocity on an incline at an angle of θ = 14.0 o , as shown above. Draw a free-body diagram for the mass. Determine numerical values for all of the forces acting on the mass. With the same vectors above we have

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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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HW06Sol - W = mg =(15.0kg(9.80m/s 2 = 147N F x = Wsin θ...

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