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

Step-by-step 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=(0.263)/(2.4)=0.109 N`

net force in Y direction

`F_y=0.160* \sin 45=0.113 N`

`a_y=(0.113)/(2.4)=0.047 m/s^2`

Thus displacement in x direction

`x=ut+(at^2)/(2)`

`x=0.15* 2.8+(0.109* 2.8^2)/(2)=0.847 m`

Displacement in Y direction

`Y=ut+(at^2)/(2)`

here u=0

`Y=0+(0.047* 2.8^2)/(2)=0.184 m`

net displacement
`√(0.847^2+0.184^2)=√(0.7513)`

=0.866 m

Direction of displacement

`tan\theta =(0.184)/(0.847)`

`\theta =12.25^(\circ)` south of east

`\theta =-12.25^(\circ)`