This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 freebody 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 ....
View
Full Document
 Spring '11
 lomant
 Force, Friction, Mass, ax

Click to edit the document details