ch05-p035 - 35. The free-body diagram is shown next. FN is...

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(a) Placing the origin at the bottom of the plane, the kinematic equations (Table 2-1) for motion along the x axis which we will use are vv a x 2 0 2 2 =+ and vv a t 0 . The block momentarily stops at its highest point, where v = 0; according to the second equation, this occurs at time tv a =− 0 . The position where it stops is () 2 2 0 2 11 ( 3 . 5 0 m / s ) 1.18 m 22 9.8 m/s sin 32.0 v x a §· ¨¸ = − ° ©¹ , or | | 1.18 m. x = (b) The time is 00 2 3.50m/s 0.674s. sin 9.8m/s sin 32.0 t ag θ == = = ° (c) That the return-speed is identical to the initial speed is to be expected since there are no dissipative forces in this problem. In order to prove this, one approach is to set x = 0 and solve xv t a t 0 1 2 2 for the total time (up and back down) t . The result is t v a v g ° = 23 5 0 98 320 135 sin . .s
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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