Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. he decides to move them at the same time by pushing on box a with a horizontal pushing force fp = 8.5 n. here, a has a mass ma = 11.0 kg and b has a mass mb = 7.0 kg. the contact force between the two boxes is fc. the coefficient of kinetic friction between the boxes and the floor is 0.04. (assume fp acts in the +x direction.)

The question is incomplete, but we can assume that the unknown in this case is the acceleration of the two boxes once acted on by the horizontal pushing force. If this is the case, the answer is a= 0.08 m/sec².

Explanation:

If the two boxes are in contact each other, we can take both boxes as it were only one with a mass equivalent to the sum of both masses, making unnecessary to find out the internal force between both boxes FC.

So, in the horizontal direction, we can apply Newton's 2nd Law, as follows:

Fnet = (mA +mB) . a

Now, in the horizontal direction, we have only 2 forces acting: FP (pushing force) and Ffr (friction force) that is aiming in the opposite direction, due to it always opposes to the relative movement between 2 surfaces, in this case, the boxes and the floor.

Assuming that the boxes are moving, the friction force is equal to the kinetic friction coefficient times the normal force.

Now, if we apply Newton's 2nd Law in the vertical direction, we conclude that the sum of external forces acting on the boxes must be 0, due to the boxes have no acceleration in the vertical direction.

In the vertical direction, we have only 2 forces acting: the one due to gravity   ((mA + mB) . g), and the normal force, that adopts any value needed to satisfy the Newton's 2nd Law, so, in this case, the normal force is numerically equal to the weight of the boxes:

N = (mA + mB) . g = 18 Kg. 9.8 m/s² = 176.4 N

So, we can write the following for Ffr:

Ffr = 0.04 * 176.4 N = 7.05 N

Fnet = 8.5 N - 7.05 N = 1.45 N⇒ a= Fnet / (mA + mB) = 1.45 N / 18 Kg

⇒ a= 0.08 m/sec²

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

a, c, d, e are all applying about success strategies.