Question

A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with magnitude of 6 000 N directed toward the northwest (45° N of W). What is the magnitude of the resultant acceleration?

Answer 1

Answer:

Magnitude of resultant acceleration = 2.21 m/s²

Explanation:

We have equation for force , F = Mass x Acceleration

A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide

3000 N = 2000 x a₁

a₁ = 1.5 m/s² eastward = 1.5 i m/s²

A wind force against its sails with magnitude of 6 000 N directed toward the northwest (45° N of W)

6000 N = 2000 x a₂

a₂ = 3 m/s² northwest (45° N of W) = -3 cos 45 i + 3 sin 45 j = -2.12 i + 2.12 j m/s²

Resultant acceleration, a = a₁+a₂ = 1.5 i -2.12 i + 2.12 j  = -0.62 i + 2.12 j

$\texttt{Magnitude of acceleration = }\sqrt{(-0.62)^2+2.12^2}=2.21 m/s^2$

Magnitude of resultant acceleration = 2.21 m/s²

Answer 2

The magnitude of the resultant acceleration of sailboat is 2.21 m/s.

The given parameter;

• mass of the sailboat, m = 2,000 kg
• eastward force experienced by the sail boat, Fₓ = 3,000 N
• northward force experienced by the sail boat, F = 6000, 45°

The horizontal component of force on the boat is calculated as;

$F_x = 3000 - 6000cos(45)\\\\F_x = -1,242.6 \ N$

The vertical component of force on the boat is calculated as;

$F_y = 6000\times sin(45)\\\\F_y = 4,242.6 \ N$

The resultant force on the boat is calculated as;

$F = \sqrt{F_x^2 + F_y^2}\\\\F = \sqrt{(-1242.6)^2 + (4242.6)^2} \\\\F = 4420.83 \ N$

The resultant acceleration of sailboat is calculated as;

F = ma

$a = \frac{F}{m} \\\\a = \frac{4420.83}{2000} \\\\a = 2.21 \ m/s^2$

Thus, the magnitude of the resultant acceleration of sailboat is 2.21 m/s.

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