A 358-kg boat is sailing 15.9° north of east at a speed of 1.80 m/s. Thirty seconds later, it is sailing 36.6° north of east at a speed of 3.68 m/s. During this time, three forces act on the boat: a 29.7-N force directed 15.9° north of east (due to an auxiliary engine), a 23.9-N force directed 15.9° 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.

To find the magnitude and direction of the force W, we need to consider the net force acting on the boat.

Net force is the vector sum of all the forces acting on an object. In this case, the net force on the boat is the sum of the forces due to the auxiliary engine, resistance due to the water, and the force W (due to the wind).

Let's break down the given forces into their x and y components:

Auxiliary engine force (F1):
Magnitude (F1) = 29.7 N
Direction (θ1) = 15.9° north of east

Breaking this force into x and y components:
F1x = F1 * cos(θ1)
F1y = F1 * sin(θ1)

Resistance force (F2):
Magnitude (F2) = 23.9 N
Direction (θ2) = 15.9° south of west

Breaking this force into x and y components:
F2x = F2 * cos(θ2)
F2y = F2 * sin(θ2)

Let's calculate the x and y components of all the forces:

F1x = 29.7 N * cos(15.9°)
F1y = 29.7 N * sin(15.9°)

F2x = 23.9 N * cos(-15.9°) (since 15.9° south of west is equivalent to -15.9° west of south)
F2y = 23.9 N * sin(-15.9°)

Now, let's calculate the x and y components of the net force:

Net force in the x-direction: Fx = F1x + F2x
Net force in the y-direction: Fy = F1y + F2y

Next, we can use the information of the velocities of the boat to determine the final velocities in the x and y directions. We will assume that the boat maintains a constant velocity between the two measurements.

Let's break down the final velocity into its x and y components:

Final velocity 1 (v1):
Magnitude (v1) = 1.80 m/s
Direction (θv1) = 15.9° north of east

Breaking this velocity into x and y components:
v1x = v1 * cos(θv1)
v1y = v1 * sin(θv1)

Final velocity 2 (v2):
Magnitude (v2) = 3.68 m/s
Direction (θv2) = 36.6° north of east

Breaking this velocity into x and y components:
v2x = v2 * cos(θv2)
v2y = v2 * sin(θv2)

Now, let's calculate the change in velocities:

Δvx = v2x - v1x
Δvy = v2y - v1y

To find the change in velocity, we can use the formula:

Δv = Δvx^2 + Δvy^2

Given the mass of the boat (m = 358 kg), we can find the magnitude of the net force:

Net force (F) = m * Δv / Δt

where Δt = 30 seconds.

Finally, to find the magnitude and direction of the force W, we can calculate its x and y components:

Wx = -Fx
Wy = -Fy

To find the magnitude of force W, we use the formula:

|W| = sqrt(Wx^2 + Wy^2)

The direction of force W can be expressed as an angle with respect to due east:

θ = arctan(Wy/Wx)

By substituting the values, you can now calculate the magnitude and direction of force W.