# ch07-p077 - t = 1.0 s Eq 7-10 gives K = K i W = 0.64 J 0.6...

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77. (a) We can easily fit the curve to a concave-downward parabola: x = 1 10 t (10 – t ), from which (by taking two derivatives) we find the acceleration to be a = –0.20 m/s 2 . The (constant) force is therefore F = ma = –0.40 N, with a corresponding work given by W = Fx = 2 50 t ( t – 10). It also follows from the x expression that v o = 1.0 m/s. This means that K i = 1 2 m v 2 = 1.0 J. Therefore, when
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Unformatted text preview: t = 1.0 s, Eq. 7-10 gives K = K i + W = 0.64 J 0.6 J ≈ , where the second significant figure is not to be taken too seriously. (b) At t = 5.0 s, the above method gives K = 0. (c) Evaluating the W = 2 50 t ( t – 10) expression at t = 5.0 s and t = 1.0 s, and subtracting, yields –0.6 J. This can also be inferred from the answers for parts (a) and (b)....
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## This note was uploaded on 03/19/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.

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