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A 10Kg monkey climbs up a massless rope that runs over a frictionless branch of a tree and back down to a 15Kg package on the ground. The magnitude of the least acceleration the monkey must have if it is to lift the package off the ground is_____?

Aditya , 5 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 5 Years ago

Dear Aditya
 
Sum of forces on monkey = MA=T-Mg Sum of forces on banana box = ma=T-mg when the package goes from being on the ground to being lifted off the ground it's acceleration will change from 0 to some number, thus the minimum rope tension required will be just greater than T when a=0 ma=T-mg a=(T-mg)/m 0=(T-mg)/m T=mg=(15 kg)(9.8 m/s^2)=147 N MA=T-Mg A=(T-Mg)/M A=[147 N - (10 kg)(9.8 m/s^2)]/10 kg = 4.9 m/s^2

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