# ch07-p074 - 74. (a) The component of the force of gravity...

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W 1 22 27 10 40 10 =− × × .. N (1.5 m) J. c h (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 W 2 45 9 8 4 0 10 == × (. . kg) m / s (0.91 m) J. c h (d) Since N F G is perpendicular to the direction of motion of the block, and cos90 ° = 0, work done by the normal force is W 3 = 0 by Eq. 7-7. (e) The resultant force G F net is zero since there is no acceleration. Thus, its work is zero, as can be checked by adding the above results WWW 123 0 ++= . 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.

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