Two blocks are in contact on a frictionless table. A horizontal force is applied

Question

Two blocks are in contact on a frictionless table. A horizontal force is applied to the larger block, as shown in the figure. (a) If m1 = 3.0 kg, m2 = 1.4 kg, and F = 3.6 N, find the magnitude of the force between the two blocks. (b) Assume instead that a force of the same magnitude F is applied to the smaller block but in the opposite direction, and calculate the magnitude of the force between the blocks. (Why is the value calculated in (b)not the same as that calculated in (a)?)

I got 2.5 N for part b but I can't figure out the answer to part a (it's not 1.1 N).

Explanation / Answer

The method is same in which you got the second part. So the correct answer to part a is 1.15 or 1.2(approx).

a) the force between the blocks is called contact force given by Fc = m2 F / m1 + m2

ie., Fc = 1.4*3.6 /4.4 = 1.15 N = 1.2 N

b) the contact force in this case is Fc = m1*F /m1 + m2 = 3*3.6/4.4 = 2.45 N = 2.5 N

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