A box of Cheerios and a box of Wheaties are accelerated accross a horizontal force vector F applied to the Cheerios box. THe magnitude of the frictional force on the Cheerios box is 2 N and the magnitude of the frictional force on the Wheaties box is 4 N. If the magnitude of the vector focre is 12 N what is the magnitude of the force on the Wheaties box from the Cheerios box?

Can you clearify what I have to do?

You need to determine the acceleration first, since you cannot apply force equilibrium to any box.

Fnet = (M1 + M2) a

Fnet = 12 - 2 - 4 = 6 N accel;erates the two boxes together.

Solve for a and then apply
F = M1 a
to the Cheerios box alone, to solve for the force net F acting on the Cheerios box.
F = 12 - F', where F' is the force applied by the Wheaties box in front of it.
F' is the force that they want you to compute

To find the magnitude of the force on the Wheaties box from the Cheerios box, we first need to calculate the acceleration (a) of the two boxes together. We do this by using the formula:

F_net = (M1 + M2) * a

In this case, the net force (F_net) is given as 12 N, and the frictional forces on the Cheerios and Wheaties boxes are 2 N and 4 N respectively. So, we have:

F_net = 12 N
frictional force on Cheerios box = 2 N
frictional force on Wheaties box = 4 N

Using the formula:

F_net = force on Cheerios box + force on Wheaties box - frictional force on Cheerios box - frictional force on Wheaties box

12 N = force on Cheerios box + force on Wheaties box - 2 N - 4 N

Simplifying the equation:

12 N + 6 N = force on Cheerios box + force on Wheaties box

18 N = force on Cheerios box + force on Wheaties box

Now, we need to solve for the force on the Cheerios box (force on Cheerios box = F' in the problem statement):

F' = 18 N - force on Wheaties box

Finally, substituting the value of the force on Wheaties box (given as 4 N):

F' = 18 N - 4 N

F' = 14 N

Therefore, the magnitude of the force on the Wheaties box from the Cheerios box is 14 N.

To solve this problem, we need to find the magnitude of the force on the Wheaties box from the Cheerios box.

First, we can calculate the acceleration of the two boxes together using Newton's second law:

Fnet = (M1 + M2) * a

Given that Fnet (the net force) is 12 N, the frictional force on the Cheerios box is 2 N, and the frictional force on the Wheaties box is 4 N, we can substitute these values into the equation:

12 N = (M1 + M2) * a

However, since we are only interested in the force on the Wheaties box from the Cheerios box, we need to find the acceleration first.

By rearranging the equation and substtituting the given values, we can solve for the acceleration:

a = (12 N - 2 N - 4 N) / (M1 + M2)

Simplifying, we get:

a = 6 N / (M1 + M2)

Now that we have the acceleration, we can calculate the force net F acting on the Cheerios box alone using the formula F = M1 * a. Rearranging this formula to solve for F, we get:

F = 12 N - F'

Where F' is the force applied by the Wheaties box in front of it.

Therefore, the magnitude of the force on the Wheaties box from the Cheerios box is the difference between the net force of 12 N and the force on the Cheerios box, F.

8.8N