ch05 - CHAPTER 5 SOLUTION FOR PROBLEM 13(a The free-body...

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C HAPTER 5S OLUTION FOR P ROBLEM 13 (a) The free-body diagram is shown in Fig. 5–18 of the text. Since the acceleration of the block is zero, the components of the Newton’s second law equation yield T mg sin θ = 0 and F N mg cos θ = 0. Solve the first equation for the tension force of the string: T = mg sin θ = (8 . 5 kg)(9 . 8m / s 2 ) sin 30 =42N . (b) Solve the second equation for F N : F N = mg cos θ =(8 . 5 kg)(9 . / s 2 ) cos 30 =72N . (c) When the string is cut it no longer exerts a force on the block and the block accelerates. The x component of the second law becomes mg sin θ = ma ,so a = g sin θ = (9 . / s 2 ) sin 30 = 4 . 9m / s 2 . The negative sign indicates the acceleration is down the plane.
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C HAPTER 5S OLUTION FOR P ROBLEM 29 The free-body diagram is shown at the right. n F N is the normal force of the plane on the block and m n g is the force of gravity on the block. Take the positive x axis to be down the plane, in the direction of the acceleration, and the positive y axis to be in the direction of the normal force. The x component of Newton’s second law is then mg sin θ = ma , so the acceleration is a = g sin θ . (a) Place the origin at the bottom of the plane. The equations for motion along the x axis are x = v 0 t + 1 2 at 2 and v = v 0 + at . The block stops when v =0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . F N mg θ According to the second equation, this is at the time t = v 0 /a . The coordinate when it stops is x = v 0 w v 0 a W + 1 2 a w v 0 a W 2 = 1 2 v 2 0 a = 1 2 v
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ch05 - CHAPTER 5 SOLUTION FOR PROBLEM 13(a The free-body...

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