A 32.7-kg block (m1) is on a horizontal surface, connected to a 6.7-kg block (m2) by a massless string as shown in the Figure. The pulley is massless and frictionless. A force of 236.1 N acts on m1 at an angle of 28.7o. The coefficient of kinetic friction between m1 and the surface is 0.245. Determine the upward acceleration of m2.

To determine the upward acceleration of m2, we need to consider the forces acting on both blocks and apply Newton's second law.

First, let's identify the forces acting on each block:

For m1:
1. The force applied at an angle of 28.7° (236.1 N)
2. The force of kinetic friction (Fk1)

For m2:
1. The force of tension in the string (T)
2. The force of gravity (m2 * g), where g is the acceleration due to gravity

Let's break down the force applied at an angle into its x (horizontal) and y (vertical) components:
- The x-component of the force applied to m1 is Fx1 = F1 * cos(θ) = 236.1 N * cos(28.7°)
- The y-component of the force applied to m1 is Fy1 = F1 * sin(θ) = 236.1 N * sin(28.7°)

Now, we can write the equations of motion for each block:

For m1:
1. ΣFx1 = T - Fk1 = m1 * ax, where ax is the acceleration of m1
2. ΣFy1 = N - m1 * g - Fy1 = 0, since m1 does not move vertically

For m2:
1. ΣFy2 = T - m2 * g = m2 * ay, where ay is the acceleration of m2

Now, let's solve for the unknowns. We'll start with m1:

1. ΣFx1 = T - Fk1 = m1 * ax
Rearranging the equation, we get:
T = Fk1 + m1 * ax

2. ΣFy1 = N - m1 * g - Fy1 = 0
Since m1 does not accelerate vertically, we can write:
N = m1 * g + Fy1

Next, let's move on to m2:

3. ΣFy2 = T - m2 * g = m2 * ay

Now, we need to determine the force of kinetic friction, Fk1:

Fk1 = μ * N
= μ * (m1 * g + Fy1)

Finally, we substitute these values back into equation (1) to solve for the acceleration of m1, ax, and equation (3) to solve for the acceleration of m2, ay:

T = Fk1 + m1 * ax
= μ * (m1 * g + Fy1) + m1 * ax

Substituting for Fy1:
T = μ * (m1 * g + F1 * sin(28.7°)) + m1 * ax

ΣFy2 = T - m2 * g = m2 * ay
T - m2 * g = m2 * ay

Now we have two equations with two unknowns. We can solve these equations simultaneously to find the values of ax and ay.

Now wait a minute. I just did one like this below. You try now. I already passed the course.