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Unformatted text preview: m g F T x y F L d 41. We choose the origin for x at the bottom of the plane. Note that down the plane (the direction of the acceleration) is positive. ( a ) From the force diagram for the block, we have ? F = m a : x-component: mg sin = ma ; y-component: F N mg cos = 0. From the x-equation we find the acceleration: a = g sin = (9.80 m/s 2 ) sin 22 = 3.67 m/s 2 . For the motion of the block, we find the distance until it momentarily stops from v 2 = v 2 + 2 a ( x x ); 0 = ( 4.0 m/s) 2 + 2(3.67 m/s 2 )( x 0), which gives x = 2.2 m. Thus the block travels 2.2 m up the plane . ( b ) We find the time from the start to return to the bottom from x = x + v t + ! at 2 ; 0 = 0 + ( 4.0 m/s) t + ! (3.67 m/s 2 ) t 2 , which gives t = 0 (the start), and...
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