ch05-p057 - 57 We take y to be up for both the monkey and...

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then F N = 0 and a p = 0. According to the second law equation for the package, this means F = m p g . Substituting m p g for F in the equation for the monkey, we solve for a m : () 2 2 15 kg 10 kg 9.8 m/s 4.9 m/s . 10 kg pm m m mm g Fm g a == = = (b) As discussed, Newton’s second law leads to pp p gm a −= for the package and m Fmgm a for the monkey. If the acceleration of the package is downward, then the acceleration of the monkey is upward, so a m = – a p . Solving the first equation for F ga m =+ =− di bg and substituting this result into the second equation, we solve for a m : 2 2 15 kg 10 kg 9.8 m/s 2.0 m/s . 15 kg 10 kg m g a = ++ (c) The result is positive, indicating that the acceleration of the monkey is upward. (d) Solving the second law equation for the package, we obtain ( ) 22 15 kg 9.8 m/s 2.0 m/s 120N. Fmga = = 57. We take + y to be up for both the monkey and the package. (a) The force the monkey pulls downward on the rope has magnitude
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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