Unformatted text preview: 76. (a) Eq. 710 (along with Eq. 71 and Eq. 77) leads to vf = (2 m F cosθ )1/2= (cosθ )1/2, where we have substituted F = 2.0 N, m = 4.0 kg and d = 1.0 m. (b) With vi = 1, those same steps lead to vf = (1 + cosθ )1/2. (c) Replacing θ with 180º – θ, and still using vi = 1, we find vf = [1 + cos(180º – θ )]1/2 = (1 – cosθ )1/2. (d) The graphs are shown on the right. Note that as θ is increased in parts (a) and (b) the force provides less and less of a positive acceleration, whereas in part (c) the force provides less and less of a deceleration (as its θ value increases). The highest curve (which slowly decreases from 1.4 to 1) is the curve for part (b); the other decreasing curve (starting at 1 and ending at 0) is for part (a). The rising curve is for part (c); it is equal to 1 where θ = 90º.
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 Spring '09
 NAJAFZADEH
 mechanics, General Relativity, Euclidean vector, highest curve

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