Question

A 350 kg sailboat has an acceleration of 0.62 m/s^2 at an angle of64 degrees north of east. Find the magnitude and direction of thenet force that acts on the sailboat.a. 0 (zero) Nb. 130 N, 64° north of eastc. 220 N, 64° north of eastd. 130 N, 64° south of weste. 220 N, 64° south of west

Answer 1

The magnitude of the net force on the sailboat is 333 N, and the direction is 65.3 degrees north of east.

Step-by-step explanation:

To find the magnitude and direction of the net force, we need to first find the horizontal and vertical components of the force. The horizontal component can be calculated using the formula F_h = F * cos(theta), and the vertical component can be calculated using the formula F_v = F * sin(theta), where F is the magnitude of the force and theta is the angle of the force.

In this case, the horizontal component of the force can be calculated as F_h = 350 kg * 0.62 m/s2 * cos(64 degrees) = 130 N, and the vertical component can be calculated as F_v = 350 kg * 0.62 m/s2 * sin(64 degrees) = 306 N.

The magnitude of the net force can be calculated using the formula F_net = sqrt(F_h2 + F_v2) = sqrt((130 N)2 + (306 N)2) = 333 N. The direction of the net force can be found using the tangent function, where tan(theta_net) = (F_v / F_h) = (306 N / 130 N) = 2.35. Taking the arctan of both sides, we find that theta_net = 65.3 degrees.

Therefore, the magnitude of the net force is 333 N and the direction is 65.3 degrees north of east.

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