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CH07.Problems

CH07.Problems. JH.131. Position  velocity  acceleration  force 2 times & position  displacement Force & displacement  work V= -10 t, a= -10, F= -10*5=-50 N; Xi= 10, Xf = 10-5*16, d = -80 m W = F . D = 4 000 J. K =1/2 m V^2, doubling V makes V^2 4 times higher, Vi  Vf =2 Vi

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CH07.Problems

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  1. CH07.Problems JH.131

  2. Position  velocity  acceleration  force • 2 times & position  displacement • Force & displacement  work • V= -10 t, a= -10, F= -10*5=-50 N; • Xi= 10, Xf= 10-5*16, d = -80 m • W = F . D = 4 000 J

  3. K =1/2 m V^2, doubling V makes V^2 4 times higher, Vi Vf=2 Vi Thus Ki Kf= 4 Ki

  4. Ws = ½ k (Xi^2 - Xf^2) • Or • Fs = -k X and Ws = - Fs * x • Negative of the integral • Take two triangles: • ½ 8 * (-80) + ½ (-4) * 40 = -240 N cm = - 2.4 J • Then W = - integral =+2.4J

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