A duck has a mass of 2.7 kg. As the duck paddles, a force of 0.14 N acts on it i

Question

A duck has a mass of 2.7 kg. As the duck paddles, a force of 0.14 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.14 N in a direction of 43° south of east. When these forces begin to act, the velocity of the duck is 0.16 m/s in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 2.1 s while the forces are acting.

I've done it a few times the way the book has it but keep getting different answers/wrong answers

Explanation / Answer

First, we need to find the total x and y forces.

There are two x forces, the .14 N east, and the x piece of .14 N south of east. To get the x component, we form a right triangle, with .14 N pointing below the x axis by 43 degrees. Then, we see that using cosine we can get the x component.

So, cos() = adjacent/hypotenuse

cos(43) = x compononet / .14 N

x component = .1024 N.

We add this to the other force which was already east at .14 N, for a total of .2424 N EAST.

Now, we get the y component by using sin, so

sin() = y component / hypotenuse

sin (43) = y /.14 N, therefore y = .0955 N SOUTH

Next, find the x and y ACCELERATION using F=ma, so:

Fx = ( 2.7 kg)ax

.2424 N = 2.7ax

ax = .0898 m/s2

Now, lets find displacment in the x direction using dx = vixT + ( 1/2)axT2

dx = ( .16 m/s)( 2.1 sec) + (1/2)(.0898 m/s2)(2.1 sec)2

dx = .534 m

Similarly, lets do y direction.

Fy = may

.0955 N = (2.7 kg) ay

ay = .0354 m/s2

And then use dy = viyT + ( 1/2)ayT2 however, initial y velocity is zero, so

dy = (1/2)(.0354 m/s2)(2.1 sec)2

dy = .078 m

We now put the dx and dy together, as the legs of a right triangle, and find the resultant hypotenuse.

.534 m 2 + .078 m 2 = c2

resultant = .54 m

direction , use tangent.

tan() = y/x

tan() = .078 / .534

= 8.31 south of east

Final answer: .54 m @ 8.31 south of east

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