Bolt from the Blue | Part I

First note that, if the $10$ newton force is applied instantaneously to the $2$ kg block, then since $F_{f} = \mu N = 0.5 * 2 * 9.8 = 9.8 \lt 10$ the $2$ kg block would briefly slide along the top of the $3$ kg block. However, since the surface below the $3$ kg block is frictionless this block would be dragged in the direction of the force as well, to the point that the two blocks will be moving together with the same acceleration.

If the force is applied more gradually then no slippage would occur, since before the applied force reached its full amount the lower block would be moving in tandem with the upper block. In this case the $10$ newton force is in essence being applied to both blocks, mass $M = 2 + 3 = 5$ kg., in which case the acceleration of the system would be

$a = \dfrac{F}{M} = \dfrac{10}{5} = 2 \frac{m}{s}$ .

Analyzing this result in more detail, look at the horizontal forces on the lower block in isolation, assuming that the two blocks are accelerating together at the same rate $a$ . The only horizontal force directly applied to this block is the force of friction, $F_{f}$ , between the two blocks, which is dragging the block to the right. This then equals the product of this block's mass by its acceleration, and so $F_{f} = 3a$ newtons.

Next we look at the horizontal forces on the upper block in isolation. There is the $10$ newton force to the right and the force of friction to the left. So

$10 - F_{f} = 2a \Longrightarrow 10 - 3a = 2a \Longrightarrow a = 2 \frac{m}{s}$ ,

in agreement with the previous result. So the result from both analyses is that $F_{f} = 3a = \boxed{6}$ newtons.