ok part (b)

A 4.0 kg block is stacked on top of a 12.0 kg block, whcih is accelerating along a horizontal table at a = 5.2 s^-2 m. Let Mu k = Mu s = Mu.

(a) What mininum coefficent of friction Mu between the two blocks will prevent the 4.0 kg block from sliding off?

(b) If Mu is only half this mininium vale, what is the acceleration of the 4.0 kg block with respect to the table, and

(c) with respect to the 12.0 kg block?

(d) What is the force that must be applied to the 12.0 kg block in (a) and in (b), assuming that the table is frictionless?

for (a) I got .66

ok for (b) I got 3.2 s^-2 m

for (c) I got 2.0 s^-2 m

And I don't understand what the problem means by

"(d) What is the force that must be applied to the 12.0 kg block in (a) and in (b), assuming that the table is frictionless?"

Does this mean that for (a) the bottom block experiances a force of kinetic friction were Mu is the same Mu i found in part(a) of .66????

part (b) is frictionless with the table... I know how to do that just don't understand what the other part is asking??????

Sense this problem 32 the answer isn't in the back =[

can you check my answer for (c)

ok for the applied force for (a) on the bottom block I got 83 N not sure if it's right...

for the applied force in (b) wouldn't it be the same... I'm confused...

Can you check my answer for (c)

The bottom block has no friction with the table.

If you know the acceleration, then the force applied must be (totalmass)a where totalmass is 12+4 kg.

In part (d) of the problem, you are asked to determine the force that must be applied to the 12.0 kg block in both scenario (a) and scenario (b), assuming that the table is frictionless.

Let's break it down:

(a) In scenario (a), where the minimum coefficient of friction (Mu) is determined to be 0.66, the 4.0 kg block is prevented from sliding off the 12.0 kg block. This means that to keep the 4.0 kg block in place, the static friction force acting between the two blocks must be at least equal to the force trying to move the 4.0 kg block. Therefore, you can calculate the static friction force using the equation:

Friction force (static) = Mu * normal force

where "normal force" is the force exerted by the 12.0 kg block on the 4.0 kg block, which is equal to its weight: normal force = m * g

Using this information, you can calculate the force applied to the 12.0 kg block, which is equal in magnitude but opposite in direction to the static friction force. Keep in mind that this force is the force required to just prevent sliding, so the acceleration of the blocks will be zero.

(b) In scenario (b), Mu is only half the minimum value found in part (a), so it is halved to 0.33. Since Mu is lower, the friction force (kinetic) between the two blocks will be reduced. In this case, the 4.0 kg block can slide on the 12.0 kg block, and the acceleration of the 4.0 kg block will be determined by the net force acting on it. Since we're assuming a frictionless table in this scenario, the only forces acting on the block are its weight and the force exerted by the 12.0 kg block.

To find the acceleration of the 4.0 kg block in scenario (b), you need to calculate the net force acting on it, which is the difference between the force exerted by the 12.0 kg block and the weight of the 4.0 kg block. Then, divide this net force by the mass of the block to find the acceleration.

(c) In part (c), you are asked to determine the acceleration of the 4.0 kg block with respect to the 12.0 kg block. This refers to the relative motion between the two blocks. Since the table is assumed to be frictionless, the acceleration of the 4.0 kg block with respect to the 12.0 kg block will be the same as the acceleration determined in part (b), as there are no other forces affecting their relative motion.

I hope this clarifies the meaning behind part (d) and helps you understand how to approach the problem. Let me know if you have any further questions!