Question

A duck has a mass of 2.40 kg. As the duck paddles, a force of 0.150 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.160 N in a direction of 45.0° south of east. When these forces begin to act, the velocity of the duck is 0.150 m/s in a direction due east. Find (a) the magnitude and (b) the direction (relative to due east) of the displacement that the duck undergoes in 2.80 s while the forces are acting. (Note that the angle will be negative in the south of east direction.)

Answer 1

Answer:

Explanation:

Given

mass of duck=2.40 kg

$F_1$=0.150 N

$F_2$=0.160 N in a direction south of east

Net force in x- direction

$F_x=0.150+0.160\cos 45=0.263 N$

$a_x=\frac{0.263}{2.4}=0.109 N$

net force in Y direction

$F_y=0.160\times \sin 45=0.113 N$

$a_y=\frac{0.113}{2.4}=0.047 m/s^2$

Thus displacement in x direction

$x=ut+\frac{at^2}{2}$

$x=0.15\times 2.8+\frac{0.109\times 2.8^2}{2}=0.847 m$

Displacement in Y direction

$Y=ut+\frac{at^2}{2}$

here u=0

$Y=0+\frac{0.047\times 2.8^2}{2}=0.184 m$

net displacement $\sqrt{0.847^2+0.184^2}=\sqrt{0.7513}$

=0.866 m

Direction of displacement

$tan\theta =\frac{0.184}{0.847}$

$\theta =12.25^{\circ}$ south of east

$\theta =-12.25^{\circ}$