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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 xx = 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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 '10
 Sauke,ToddB
 Physics, Force, Mass, #, 15 minutes, 45 kg

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