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Physics_50_Prof_Test_3s_key

# Physics_50_Prof_Test_3s_key - Σ F y = n – mg cos θ = 0...

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Physics 50 Proficiency Test 3.s (Time: 15 minutes) 25 points By Todd Sauke Name ___________ KEY ________ Section # __ KEY ___ A sled (of total mass 45 kg including passenger) starts from rest at the top of a long snowy hill of constant 25º incline to the horizontal. The coefficient of kinetic friction, μ k , between the sled's runners and the snow is 0.05. The sled accelerates straight down the slope and we wish to determine its speed after having traveled 110 meters. You should use 9.8 m/s 2 as the magnitude of the acceleration due to gravity. Make a free body diagram below, incorporating all of the features that a good free body diagram should have, including appropriately oriented axes with positive direction shown, and showing all of the forces acting on the sled. Solve the "y axis" (normal to the surface) equilibrium problem to determine the magnitude of the normal force, n.

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Unformatted text preview: Σ F y = n – mg cos( θ ) = 0 n = mg cos( θ ) = 45 (9.8) cos(25º) = 400 N n = _______ 400 (400) ______N Solve the "x axis" (tangential to the surface) problem to determine the acceleration down the hill. Σ F x = mg sin( θ ) – f = mg sin( θ ) – μ k n = m a x m g sin( θ ) – m μ k g cos( θ ) = m a x a x = g ( sin( θ ) – μ k cos( θ ) ) = 9.8 x ( 0.42 – 0.05 x .91 ) = 3.67 a = ______ 3. 67 ___ (3.7) _______m/s 2 Using one of the listed kinematic equations, find the speed of the sled at the bottom of the hill, 110 m from the starting point. v = v 0 + a t x-x = ½(v +v) t v 2 = v 2 + 2a ( x – x ) x = x 0 + v t + ½ at 2 v 2 = v 2 + 2 a x (x – x o ) = 2 a x (110) __________ ________________ v = √ 2 a x (110) = √ 2 x 3.67 x 110 = 28.4 v = _______ 28.4 ____ (28) ____m/s...
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