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three blocks in contract with each other are pushed across

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asked Sep 26, 2014 in AP Physics C by ang2014 (410 points)

1 Answer

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Three blocks in contact with each other are pushed across a rough horizontal surface by a 110 N force as shown. The acceleration of gravity is 9.8 m/s2 . If the coefficient of kinetic friction between each of the blocks and the surface is 0.14, find the magnitude of the force exerted on the 3.1 kg block by the 2.6 kg block. 

First, treat the system as one unit with a total mass ||M=2+3.1+2.6 =7.7kg|| and find its acceleration. The total normal force is ||N=Mg||. Then the total frictional force is ||f=\mu N||. The net force is ||110-f=Ma||, which gives ||a|| and all blocks accelerate with the same ||a||.

From Newton's Third Law, the force that the 2.6kg block exerts on the 3.1 kg block is equal and opposite to the force the 3.1kg block applies on the 2.6kg block. This latter force (call it ||F_3||) makes the 2.6kg block accelerate at the ||a|| found above. Then we have ||F_3-f=2.6a||. Here ||f=\mu 2.6g||.  ||F_3|| can now be found. 

answered Sep 26, 2014 by PhysicsDude (21,750 points)
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