A duck has a mass of 2.9 kg. As the duck paddles, a force of 0.15 N acts on it i

Question

A duck has a mass of 2.9 kg. As the duck paddles, a force of 0.15 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.28 N in a direction of 60° south of east. When these forces begin to act, the velocity of the duck is 0.13 m/s in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 3.7 s while the forces are acting.

Explanation / Answer

Greetings... For F1: x=0.15N , y=0 F1 =0.15 i N For F2: x is 0.28cos(-60)=0.14, y is 0.28sin(-60)= -0.243 F2 = 0.14 i N -0.243 j N Here i, j are unit vectors along x and y axes respectively. Net force F = F1 + F2 = 0.15 iN + 0.14 i N -0.243 j N --> F = (0.29 i - 0.243j)N Mass M = 2.9 kg Acceleration a = F/M Or a = (0.1 i - 0.084 j)m/s^2 Initial velocity v0 = 0.13 m/s i Time t = 3.7 s Displacement S = v0 t + 1/2 at^2 = 0.13 * 3.7 i + 1/2*(0.1 i - 0.084 j)*3.7^2 = 0.481 i + 0.69 i - 0.58 j = 1.171 i - 0.58 j Magnitude of S = sqrt(1.171^2 + 0.58^2) = 1.708 m Direction = atan(0.58/1.171) south of east = 18.8 deg south of east. Best Regards!

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