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Unformatted text preview: 109. We resolve this appropriate components. horizontal force into (a) Applying Newton’s second law to the x (directed uphill) and y (directed away from the incline surface) axes, we obtain F cosθ − f k − mg sinθ = ma FN − F sinθ − mg cosθ = 0. Using fk = μk FN, these equations lead to a= F (cosθ − μk sinθ ) − g (sinθ + μk cosθ ) m which yields a = –2.1 m/s2, or a  = 2.1 m/s2 , for μk = 0.30, F = 50 N and m = 5.0 kg. (b) The direction of a is down the plane. (c) With v0 = +4.0 m/s and v = 0, Eq. 216 gives Δx = − (4.0 m/s) 2 = 3.9 m. 2(−2.1 m/s 2 ) (d) We expect μs ≥ μk; otherwise, an object started into motion would immediately start decelerating (before it gained any speed)! In the minimal expectation case, where μs = 0.30, the maximum possible (downhill) static friction is, using Eq. 61, f s , max = μ s FN = μ s (F sinθ + mg cosθ ) which turns out to be 21 N. But in order to have no acceleration along the x axis, we must have f s = F cosθ − mg sinθ = 10 N (the fact that this is positive reinforces our suspicion that f s points downhill). Since the fs needed to remain at rest is less than fs,max then it stays at that location. ...
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Force

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