Question

A 325-g model boat facing east floats on a pond. The wind in its sail provides a force of 1.85 N that points 25° north of east. The force on its keel is 0.782 N pointing south. The drag force of the water on the boat is 0.750 N toward the west. If the boat starts from rest and heads east, how fast is it moving after it travels for a distance of 3.55 m?

Answer 1

Answer:

Vf = 4.40  m/s  and  θ = 88º

Explanation:

To solve this problem, let's look for the resultants of the force and with this we calculate the accelerations in each axis.

Let's use trigonometry to break down the forces

     Sin 25 = F1x / F1

     Cos 25 = F1y / F1

     Fix = F1 sin 25

     F1x = 1.85 sin 25

     F1x = 0.78 N

     F1y = 1.85 cos 25

     F1y = 1.67 N

     F2 = - 0.782 N j ^  (south)

     F3 = - 0.750 N i ^    (west)

We write Newton's second law

X axis (East-West)

     F1x - F3 = m ax

     ax = (F1x - F3) / m

     ax = (0.78 - (0.750)) / 0.325

     ax = 0.092 m / s²

Y axis (North-South)

     F1y - F2 = m ay

     ay = (1.67- (0.782)) / 0.325

     ay = 2.73 m / s²

Let's calculate the magnitude and direction of the acceleration

     a = RA ax2 + ay2

     a = RA 0.092² + 2.73²

     a =  2.73m / s²

     tan θ = ay / ax

     θ = tan⁻¹ (2.73/0.092)

     θ = tan⁻¹ 29.67

     θ = 88º

We calculate the speed, notice that we use the total acceleration to be able to use the totol displacement

Vf² = vo² + 2 at D

Vf² = 0 + 2 2.73 3.55

Vf = √ 19.38

Vf = 4.40  m / s

θ = 88º