A block of mass m = 2.00 kg rests on the left edge of a block of length L = 3.00

Question

A block of mass m = 2.00 kg rests on the left edge of a block of length L = 3.00 m and mass M = 8.00 kg. The coefficient of kinetic friction between the two blocks is uk = 0.300, and the surface on which the 8.00 kg block rests is frictionless. A constant horizontal force of magnitude F = 12.0 N is applied to the 2.00 kg block. How long will it take before this block makes it to the right side of the 8.00 kg block? (Note: Both blocks are set in motion when F is applied.)
I have been trying this problem for hours now, and nothing seems to work. Please help.

Explanation / Answer

So for the 2 kg block, the two forces acting on it in the x-direction are the applied force and friction (in opposite directions). For the 8 kg block, an equal and opposite friction force is present. So, we shall say that the blocks move right, and so the applied force is applied rightward while the friction force on the 2 kg block points left, and the friction force on the 8 kg block also points rightward.

1. F - Ff = 2a1

2. Ff = 8a2

where F is the applied force, Ff is the frictional force, a1 is the 2 kg block's acceleration and a2 is the 8 kg block's acceleration. Ff = 2g(0.3) for BOTH blocks because it is EQUAL AND OPPOSITE for the 8 kg block by Newton's 3rd Law. solve for a1 and a2 and use them in the kinematics equation x = vit + 0.5at^2 where x1 = x2 + 3. Now you have two equations with two variables. Just solve for t now!!

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