A 343-kg boat is sailing 16.2° north of east at a speed of 1.98 m/s. Thirty seconds later, it is sailing 37.5° north of east at a speed of 3.76 m/s. During this time, three forces act on the boat: a 30.3-N force directed 16.2° north of east (due to an auxiliary engine), a 22.3-N force directed 16.2° south of west (resistance due to the water), and W (due to the wind). Find the magnitude and direction of the force W. Express the direction as an angle with respect to due east.

Compute the velocity change vector during the 30 seconds and divide it by 30 for the average acceleration vector.

Multiply that by the mass for the net force vector, Fnet. You know two of the three vectors that make up Fnet. Use that to determine the third (Wind) force vector

To find the magnitude and direction of the force W, we need to analyze the net forces acting on the boat. We can use Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration:

Net Force = mass × acceleration

From the problem, we know the mass of the boat is 343 kg. To calculate the acceleration, we need to consider the change in velocity over time. Using the change in velocity, we can calculate the average acceleration experienced by the boat during the time interval.

First, let's find the change in velocity in the eastward direction:

ΔvE = v2E - v1E
ΔvE = (3.76 m/s)cos(37.5°) - (1.98 m/s)cos(16.2°)
ΔvE = 3.0767 - 1.9238 m/s
ΔvE = 1.1529 m/s

Next, let's find the change in velocity in the northward direction:

ΔvN = v2N - v1N
ΔvN = (3.76 m/s)sin(37.5°) - (1.98 m/s)sin(16.2°)
ΔvN = 2.2362 - 0.5684 m/s
ΔvN = 1.6678 m/s

Now, we can calculate the average acceleration:

aE = ΔvE / Δt
aE = (1.1529 m/s) / (30 s)
aE = 0.03843 m/s²

aN = ΔvN / Δt
aN = (1.6678 m/s) / (30 s)
aN = 0.05559 m/s²

Next, let's consider the forces acting on the boat. The given forces are:

F1 = 30.3 N, direction: 16.2° north of east
F2 = 22.3 N, direction: 16.2° south of west

We need to resolve these forces into their eastward and northward components to determine their effects on the boat's motion.

Resolving F1:

F1E = F1 × cos(16.2°)
F1E = 30.3 N × cos(16.2°)
F1E ≈ 28.640 N

F1N = F1 × sin(16.2°)
F1N = 30.3 N × sin(16.2°)
F1N ≈ 8.011 N

Resolving F2:

F2E = F2 × cos(16.2°)
F2E = 22.3 N × cos(16.2°)
F2E ≈ 21.009 N

F2N = F2 × sin(16.2°)
F2N = 22.3 N × sin(16.2°)
F2N ≈ 6.089 N

Now, let's find the net forces in the eastward and northward directions:

Net F_E = F1E + F2E
Net F_E = 28.640 N + 21.009 N
Net F_E ≈ 49.649 N

Net F_N = F1N - F2N
Net F_N = 8.011 N - 6.089 N
Net F_N ≈ 1.922 N

Using the net forces in the eastward and northward directions, we can find the net force vector, Net F, using the Pythagorean theorem:

|Net F| = √(Net F_E² + Net F_N²)
|Net F| = √((49.649 N)² + (1.922 N)²)
|Net F| ≈ √(2465.480 N² + 3.693 N²)
|Net F| ≈ √2469.949 N²
|Net F| ≈ 49.699 N

Finally, we can find the direction of the force W using the inverse tangent function:

θ = arctan(Net F_N / Net F_E)
θ = arctan(1.922 N / 49.649 N)
θ ≈ arctan(0.03874)
θ ≈ 2.212°

The magnitude of the force W is approximately 49.699 N, and its direction is approximately 2.212° north of east.