problem05_90

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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5.90: Since the larger block (the trailing block) has the larger coefficient of friction, it will need to be pulled down the plane; i.e. , the larger block will not move faster than the smaller block, and the blocks will have the same acceleration. For the smaller block, , kg) 00 . 4 ( N 11.11 or , kg) 00 . 4 ( ) 30 cos ) 25 . 0 ( (sin30 kg) 00 . 4 ( a T a T g = - = - ° - ° and similarly for the larger, , kg) 00 . 8 ( N 44 . 15 a T = + a) Adding these two relations, 2 s m 21 . 2 , kg) (12.00 N 55 . 26 = = a a (note that an extra figure was kept in the intermediate calculation to avoid roundoff error).
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Unformatted text preview: b) Substitution into either of the above relations gives N. 27 . 2 = T Equivalently, dividing the second relation by 2 and subtracting from the first gives , N 11 . 11 2 N 44 . 15 2 3-= T giving the same result. c) The string will be slack. The 4.00-kg block will have 2 s m 78 . 2 = a and the 8.00-kg block will have , s m 93 . 1 2 = a until the 4.00-kg block overtakes the 8.00-kg block and collides with it....
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## This document was uploaded on 02/04/2008.

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