What is the Net Force on a Helicopter at 5 Seconds?

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SUMMARY

The net force on a helicopter at 5 seconds can be determined using the position function r=(0.020m/s^2)t^3i + (2.2m/s)tj - (0.060m/s^2)t^2k. By substituting t=5.0s into the position equation, the components of the position vector are calculated. The acceleration is derived from the second derivative of the position function, leading to the application of Newton's second law, F=ma, to find the net force vector components Fx, Fy, and Fz. This method eliminates circular calculations and provides a clear path to the solution.

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  • Understanding of vector calculus
  • Familiarity with Newton's laws of motion
  • Knowledge of derivatives and their application in physics
  • Basic grasp of kinematics and dynamics
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  • Study vector calculus to enhance understanding of position, velocity, and acceleration vectors
  • Learn about the application of Newton's laws in dynamic systems
  • Explore the concept of derivatives in physics, particularly in motion equations
  • Review kinematic equations and their derivations for various motion scenarios
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Physics students, aerospace engineers, and anyone interested in understanding the dynamics of helicopter motion and force calculations.

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The position of a 2.75x10^5N helicopter under test is given by

r=(0.020m/s^2)t^3i + (2.2m/s)tj - (0.060m/s^2)t^2k

Find the net force of the helicopter at t=5.0s express the vector F in the form Fx, Fy, and Fz

so far i plugged in 5.0 to each component in the r equation which gave me different masses: 2.5, 11.0, and 1.5

then i divided each mass by 5.0 to get velocity, then i plugged everything into the average acceleration formula to get different accelerations, and then i finally used F=ma to plug into get different forces for Fx Fy and Fz
.. and it looks like I am just going in a huge circle. not sure how to do this.
 
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