A duck, some Newtons, and other fun stuff

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SUMMARY

The discussion centers on calculating the displacement of a duck with a mass of 2.9 kg under the influence of two forces: 0.11 N due east and 0.19 N at 57° south of east. Participants emphasize the importance of determining the net force and applying Newton's second law to find acceleration. The correct approach involves calculating the change in velocity in both the x and y directions to ultimately derive the displacement vector over a time interval of 2.0 seconds. The final goal is to express the displacement's magnitude and direction relative to due east.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Basic vector addition and decomposition
  • Trigonometric functions (sine, cosine, tangent)
  • Knowledge of displacement and velocity concepts
NEXT STEPS
  • Calculate net force using vector components
  • Apply Newton's second law to find acceleration
  • Determine change in velocity for both x and y components
  • Use kinematic equations to find displacement
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Students studying physics, particularly those focusing on mechanics and vector analysis, as well as educators seeking to clarify concepts related to forces and motion.

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A duck has a mass of 2.9 kg. As the duck paddles, a force of 0.11 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.19 N in a direction of 57° 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 2.0 s while the forces are acting.

What I need are the
magnitude m
direction ° south of east

I've tried drawing free body diagrams, but seem to get stuck. I'm not quite sure how to tell where the duck will be and how to find that in degrees. This is due at 11:30 tonight EST any any guidence would be greatly appreciated!
 
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The first step is to find the net force on the duck. Then apply Newton's 2nd law.
 
I've done this thus far:

Rx = .19(cos 57) +.11(cos0) = .21348N
Ry = .19(sin 57) + .11 (sin 0) = .1593N
Ax = .21348/2.9 = .07361 Ay = .1593/2.9 = .0549

then I do the tan-1 (.0549/.07361) which yields the wrong answer.

Any suggestions?
 
CelesteSeven said:
I've done this thus far:

Rx = .19(cos 57) +.11(cos0) = .21348N
Ry = .19(sin 57) + .11 (sin 0) = .1593N
OK. (But realize that the 0.19 N force is south of east.)
Ax = .21348/2.9 = .07361 Ay = .1593/2.9 = .0549

then I do the tan-1 (.0549/.07361) which yields the wrong answer.
You don't need the angle that the acceleration makes, you need the displacement. Find the change in the velocity in each direction and use it to find the displacement from the initial position. Then worry about finding the angle.
 

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