W12227 1040 10=−××..N (1.5 m)J.ch(c) Since the displacement has a vertically downward component of magnitude 0.91 m (in the same direction as the force of gravity), we find the work done by gravity to be W2459 84 0 10==×(..kg) m / s (0.91 m)J.ch(d) Since NFGis perpendicular to the direction of motion of the block, and cos90°= 0, work done by the normal force is W3= 0 by Eq. 7-7. (e) The resultant force GFnetis zero since there is no acceleration. Thus, its work is zero, as can be checked by adding the above results WWW1230++=.74. (a) The component of the force of gravity exerted on the ice block (of mass
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This note was uploaded on 03/19/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.