An applied force of 50 N at an angle of 47.5 degrees up of East is applied to a 7.25 kg crate. Find the coefficient of friction if acceleration is 3.0 m/s/s E.

a. 0.28
b. 0.42 (I know this one is wrong)
c. 0.56
d. 0.35
e. 0.49

To find the coefficient of friction in this scenario, we will use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F = ma).

First, let's analyze the forces acting on the crate:
1. The applied force of 50 N at an angle of 47.5 degrees up East.
2. The force of friction opposing the motion.

Since we are given the acceleration (3.0 m/s²) and the mass of the crate (7.25 kg), we can calculate the net force acting on the crate using Newton's second law of motion:

F = ma
F = 7.25 kg * 3.0 m/s²
F = 21.75 N

To determine the component of the applied force acting in the horizontal direction (parallel to the motion), we need to find the cosine of the angle between the applied force and the horizontal direction:

cos(47.5°) = adjacent / hypotenuse
cos(47.5°) = F(horizontal) / 50 N

Solving for F(horizontal):
F(horizontal) = cos(47.5°) * 50 N
F(horizontal) ≈ 34.04 N

Now, we can find the force of friction by subtracting the applied force component from the net force:
Force of friction = net force - F(horizontal)
Force of friction = 21.75 N - 34.04 N
Force of friction ≈ -12.29 N

Note: The negative sign indicates that the force of friction is opposing the motion.

Finally, we can find the coefficient of friction using the equation frictional force = coefficient of friction * normal force. Since the normal force is equal to the weight of the crate (mg), we can express the equation as:

|m|g = μ |m|g
|μ| = |m| / |m|
μ ≈ 12.29 N / (7.25 kg * 9.8 m/s²)
μ ≈ 0.191

Since the coefficient of friction is positive and given with only two significant figures, the answer is approximately 0.19.

Therefore, none of the given choices (a, b, c, d, e) match the correct answer.