Correct Answer to Question 2

Our strategy is this: first, we consider a "system" of the two blocks together and find that system's acceleration.

For our system, the tension in the cord connecting the blocks is an internal force holding the system together, so it won't appear in the force equation.

The force diagram shows al of the external forces acting on our system: F itself, the weight W (of 6+2=8 kg mass) and the normal force R due to the horizontal surface. R and W are equal and opposite, so they add to zero.

The resulting force is just F (=40 N). The (total) mass M is 8 kg. So, =Ma gives 40 = 8 * a or a = 5 m/s^2.

To find the tension T in the connecting cord, we consider a new system for which T is an external force. We could choose either block; let's take the 6 kg one.

Its force diagram includes its weight W' and its normal force R'; they are equal and opposite and add to zero.

The resultant force is just T. So = M a gives T = 6*5 = 30 N.

(For practice, you should try using the 2 kg block as your system; T had better be 30 N, but directed toward the left.)

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