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A sailboat has a mass of 1.20 ✕ 103 kg and is acted on by a force of 2.90 ✕ 103 N toward the east, while the wind acts behind the sails with a force of 3.90 ✕ 103 N in a direction 45° north of east. Find the magnitude (in m/s2) and direction (in degrees north of east) of the resulting acceleration.

User KFichter
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2 Answers

4 votes

Final answer:

The magnitude of the resulting acceleration is 2.75 m/s² and the direction is 47.4° north of east.

Step-by-step explanation:

To find the magnitude and direction of the resulting acceleration of the sailboat, we need to find the net force acting on it. We can do this by adding the horizontal components of the two forces acting on the sailboat.

The force acting eastward is 2.00 × 10³ N and the force acting 45° north of east is 3.00 × 10³ N. We can resolve the force acting 45° north of east into its horizontal and vertical components to find the horizontal component. The horizontal component of the force is = 3.00 × 10³ N * cos(45°) = 3.00 × 10³ N * √(2)/2 = 3.00 × 10³ N * 0.7071 = 2.12 × 10³ N.

Adding the horizontal components of the forces, we get 2.00 × 10³ N + 2.12 × 10³ N = 4.12 × 10³ N.

The resulting acceleration can be found using Newton's second law, F = ma, where F is the net force and m is the mass of the sailboat. Rearranging the formula, we get a = F/m, where a is the acceleration.

Substituting the values, we get a = 4.12 × 10³ N / 1.50 × 10³ kg = 2.75 m/s²

So, the magnitude of the resulting acceleration is 2.75 m/s². To find the direction, we can use the tangent of the angle of the resulting force with the east direction. The angle can be found using the inverse tangent function, tan⁻¹(2.12 × 10³ N / 2.00 × 10³ N) = tan⁻¹(1.06) = 47.4°.

Therefore, the direction of the resulting acceleration is 47.4° north of east.

User Alhuck
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5.7k points
4 votes

Answer:

a = 4.04 m/s2


53.44^onorth of east

Step-by-step explanation:

The mass of the boat is m = 1200 kg

The forces acting due to wind on the boat is Fwind = 3900 N j^ (north)

and The forces acting due to water on the boat is Fwater = 2900 N i^ (east)

The net force on the boat is F = 2900 N i^ + 3900 N j^

The horizontal component of the net acceleration is ax = Fwater/m = 2.41 m/s2

vertical component of the net acceleration is ay = Fwind/m = 3.25 m/s2

Net acceleration is
a = √([(2.41)^2 + (3.25)^2])

a = 4.04 m/s2

The direction of the net acceleration is
\theta = tan^(-1)(3.25)/(2.41)=

=
53.44^onorth of east

User Miss Skooter
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5.8k points