M1 and M2 are two masses connected as shown. The pulley is light (massless) and frictionless. Find the mass M1, given that M2 (7.0 kg) accelerates downwards at 3.39 m/s2, θ is 30o, and μk is 0.45.

without the picture, I have no idea.

To find the mass M1, we can use Newton's second law of motion and the concept of friction.

First, let's draw a free-body diagram for each mass:

For M1:
- There is tension force T acting upwards.
- There is a gravitational force mg acting downwards.
- There is a frictional force fk opposing the direction of motion.

For M2:
- There is tension force T acting upwards.
- There is a gravitational force mg acting downwards.

Since the pulley is assumed to be frictionless, we do not consider any torque effects. Also, since the pulley is massless, there is no vertical acceleration for M2.

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

For M1:
1. Sum of forces in the vertical direction: T - mg = 0 (Equation 1)
2. Sum of forces in the horizontal direction: fk = T (Equation 2)

For M2:
1. Sum of forces in the vertical direction: T - mg = -m2 * a (Equation 3)

We need to find the mass M1, so we will focus on Equation 1 and Equation 2.

Equation 1: T - mg = 0
Since the angle θ is given, we can write mg = m1 * g * sin(θ).

Equation 2: fk = T
The frictional force fk = μk * N, where N is the normal force. In this case, the normal force is equal to the weight of M1, so N = m1 * g.

Substituting these expressions into Equation 2 and rearranging, we get:
μk * m1 * g = T
μk * m1 * g = mg

Looking at these two equations, we can see that T is present in both equations. Therefore, equating them we get:
μk * m1 * g = m1 * g * sin(θ)

Now, we can cancel out the common factor of g:
μk * m1 = m1 * sin(θ)

Finally, we can solve for M1:
μk = sin(θ)
m1 = M2 * μk / sin(θ)

Substituting the given values:
m1 = 7.0 kg * 0.45 / sin(30°)

Simplifying, we find:
m1 ≈ 7.62 kg

Therefore, the mass M1 is approximately 7.62 kg.