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The surface is frictionless,Two blocks in contact with each other move
to the right across a horizontal surface by the two forces shown

Determine the magnitude of the force exerted on the block with mass 1.4 kg by the block with mass 6.4 kg.

Answer in units of N

The surface is frictionlessTwo blocks in contact with each other moveto the right across a horizontal surface by the two forces shownDetermine the magnitude of class=

Respuesta :

sqdancefan

9514 1404 393

Answer:

  55 8/39 N ≈ 55 N

Step-by-step explanation:

The net force on the 1.4 kg block is proportional to its mass. That is, the ratio of net force on that block to the net force on the total mass is the same as the ratio of the block's mass to the total mass.

  net force on 1.4 kg = (1.4 kg)/(1.4 +6.4 kg) × (64 N -15 N) = (7/39)(49 N)

  net force on 1.4 kg = 8 31/39 N

Then the force applied by the 6.4 kg mass is ...

  64 N - 8 31/39 N = 55 8/39 N ≈ 55 N . . . . applied by 6.4 kg mass

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