Posted by Stan on Wednesday, March 3, 2010 at 1:26am.
Figure 624 shows an initially stationary block of mass m on a floor. A force of magnitude 0.540mg is then applied at upward angle 16 degrees. What is the magnitude of the acceleration of the block across the floor if (a)Us=.6 and Uk=.5 and (b) Us=.4 and Uk=.3?

Physics  drwls, Wednesday, March 3, 2010 at 7:04am
Is the 0.540mg 0.540 milligrams or 0.54m * g? I will assume the latter.
For case (a), the maximum possible static friction force is
Fs = (m*g cos16*  0.54m*g sin16)*Us
= 0.487 m g
The horizontal component of applied force is 0.540m*g*cos 16 = 0.519 m*g. Since this exceeds the maximum possible static friction force, the block slips and accelerates at a rate determined by
kinetic friction.
The friction force in the horizontal direction must be recomputed and subtracted from the applied force component in that direction.
For case (b), proceed similarly
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