ch05-p035 - s 2(9 8 m s 2 sin 32 ◦ ´ = − 1 18 m(b The...

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35 The free-body diagram is shown at the right. F N is the normal force of the plane on the block and mg 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 v 0 a + 1 2 a v 0 a 2 = 1 2 v 2 0 a = 1 2 v 2 0 g sin θ = 1 2 ( 3 . 50 m / s)
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Unformatted text preview: / s) 2 (9 . 8 m / s 2 ) sin 32 . ◦ ´ = − 1 . 18 m . (b) The time is t = − v a = − v g sin θ = − − 3 . 50 m / s (9 . 8 m / s 2 ) sin 32 . ◦ = 0 . 674 s . (c) Now set x = 0 and solve x = v t + 1 2 at 2 for t . The result is t = − 2 v a = − 2 v g sin θ = − 2( − 3 . 50 m / s) (9 . 8 m / s 2 ) sin 32 . ◦ = 1 . 35 s . The velocity is v = v + at = v + gt sin θ = − 3 . 50 m / s + (9 . 8 m / s 2 )(1 . 35 s) sin 32 ◦ = 3 . 50 m / s , as expected since there is no friction. The velocity is down the plane....
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