How Does Air Resistance Impact Force During an Egg Drop Project?

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SUMMARY

The discussion centers on calculating the force experienced by an egg drop vehicle weighing 0.454 kg, dropped from a height of 12.19 meters. The calculated gravitational force is 4.44 Newtons, derived from the formula 0.454 kg * 9.8 m/s². Participants clarify that during freefall, the force remains constant at this value, and there is no additional force acting on the egg or vehicle until impact. Consequently, the force cannot drop below 4.44 Newtons during the fall.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Basic knowledge of gravitational force calculations
  • Familiarity with freefall dynamics
  • Ability to perform unit conversions and calculations
NEXT STEPS
  • Research the concept of terminal velocity in freefall scenarios
  • Explore the physics of impact forces and energy transfer
  • Learn about air resistance and its effects on falling objects
  • Investigate methods to reduce impact forces in egg drop designs
USEFUL FOR

Students, educators, and physics enthusiasts involved in experimental physics, particularly those working on projects related to freefall and impact force calculations.

lee132
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For an egg drop project, I have to calculate the time interval for which the force won't exceed 3 Newtons. The egg is dropped from a height of 12.19 meters, and the vehicle that carries it weights .454 kg. Calculating the force, I get .454kg*9.8 m/s^2m which equates to 4.44 Newtons. Is the 4.44 Newtons the force of the egg drop vehicle at any moment in its freefall? Because if it is, then wouldn't that mean the force could never be equal to 3 Newtons. Or is the force increasing over the time of the fall. Thanks for your help.
 
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Hi lee132! Welcome to PF! :smile:
lee132 said:
For an egg drop project, I have to calculate the time interval for which the force won't exceed 3 Newtons. The egg is dropped from a height of 12.19 meters, and the vehicle that carries it weights .454 kg. Calculating the force, I get .454kg*9.8 m/s^2m which equates to 4.44 Newtons. Is the 4.44 Newtons the force of the egg drop vehicle at any moment in its freefall? Because if it is, then wouldn't that mean the force could never be equal to 3 Newtons. Or is the force increasing over the time of the fall. Thanks for your help.

Sorry, but I don't understand the question at all …

while the vehicle is falling, there is no force between the egg and the vehicle, and the force on the vehicle (or on the egg) is, as you say, always the same. :redface:

What is the exact question? :confused:
 

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