A cord passes over a weightless and frictionless pulley. Masses of 200 g and 300 g are attached to the ends of the cord. Find the distance the masses will move during 5th second after they are started from rest?


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Moo C. answered
This is coming from  a16 year old but I think I'm right  "This problem is a Newton's 2nd problem: SF=m*a where SF is the sum of
the external forces acting on the system, m is the total mass
accelerating and a is the rate of acceleration.

In this case the sum of the external forces is equal to the difference
in the weights of the two masses: SF=m1*g-m2*g=(m1-m2)*g while the total
mass accelerating is m1+m2. Therefore, the acceleration of this system
is: A=(m1+m2)/[(m1-m2)*g]" therefore  a=(200+300)/[(200/300)*9.8]=

76.5306122, so the distance traveled in 5 seconds is

76.5306122 + 153.061224 + 229.591837+306.122449+382.653061 which equals 1 147.95918 meters so in total that's how far the masses will move

source: en.allexperts.com

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