Forces involved in egg-drop experiment: 25 Newtons to Break an Egg?

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SUMMARY

The egg drop experiment discussed reveals that approximately 25 Newtons of force is required to break an eggshell, yet the experiment recorded only 6.59 Newtons of force upon impact, indicating the egg did not break. The calculations involved determining velocity (4.55 m/s) and acceleration (6.89 m/s²) using the equations of motion. The discussion emphasizes the importance of using appropriate kinematic equations, particularly the SUVAT equations, to accurately assess the dynamics of falling objects.

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  • Understanding of Newton's laws of motion
  • Familiarity with kinematic equations (SUVAT)
  • Basic knowledge of force and acceleration calculations
  • Ability to perform unit conversions and apply them in physics problems
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  • Study the SUVAT equations in detail for better understanding of motion
  • Learn how to calculate impact forces in physics experiments
  • Explore the relationship between mass, acceleration, and force
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Students conducting physics experiments, educators teaching kinematics, and anyone interested in understanding the forces involved in impact scenarios.

Yellowkies_3275
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Homework Statement


So I've spent many hours on here and deep deeeeeep in my textbooks trying to learn things...that pertain to the Egg drop experiment

I am almost done with everything I need but I was wondering. I've read that it takes approximately 25 Newtons of force to break an eggshell...But unless I did the math wrong (I probably did the math wrong) my project got 7 Newtons of force when we did the egg drop (It didn't break but even so, to me this number seemed, scarily low) I'm not that familiar with Newton's and I was just wondering, Is dropping an egg (with protection) and getting an answer of only 6 Newtons of force produced, logical in any way shape or form...or have I just failed me and my group?

Homework Equations

The Attempt at a Solution


This was our work:

v= 3m / 0.66 s

|

\/

v= 4.55 m/s
a= Δv/Δt or a= (vf-vi)/(tf-ti)vf= 4.55 m/s

vi=0 m/s

tf= 0.66 seconds

ti= 0 secondsa= (vf-vi)/(tf-ti)

|

\/

a=(4.55-0)/(0.66-0)

|

\/

a=4.55/0.66

|

\/

a= 6.89 m/s^2

Mass of Average Egg: 0.057 kg

Mass of Parcel: 0.90 kg

Combined Mass: 0.957 kgVelocity: 4.55 m/s

ΔV: 4.55 m/s

ΔT: 0.66 s

Acceleration: 6.89 m/s^2

Mass: 0.957 kg

ΣF: 6.59 Np=mvΔp=m*Δv → Δp=(0.957)*(4.55) Δv=a*Δt → Δv= (6.89)*(0.66)Δp=m*a*Δt m*a=ΣFΔp=ΣF*Δt → Δp= (6.59)*(0.66)Δp= 4.35 kg*m/s

also since this is for a grade please don't give me answers since I don't do plagiarism and this is my work and I want to keep it as authentic as possible.I just want to know If my numbers are within reasonable range
 
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Can you talk us through your working...
Yellowkies_3275 said:
v= 3m / 0.66 s
|
\/
v= 4.55 m/s

Are you saying the egg took 0.66 seconds to fall 3m?
 
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CWatters said:
Can you talk us through your working...Are you saying the egg took 0.66 seconds to fall 3m?
yes... i timed it and every time that was around the number i got, plus we got it on video and on video it didnt take more than a second either
 
Yellowkies_3275 said:
yes... i timed it and every time that was around the number i got, plus we got it on video and on video it didnt take more than a second either
Eggs do not normally break while they are falling. They break while they are stopping.
 
Yellowkies_3275 said:
yes... i timed it and every time that was around the number i got, plus we got it on video and on video it didnt take more than a second either
Ok but you can't calculate the impact velocity like that because the egg is accelerating not falling at a constant velocity. You need to use the equations of motion or SUVAT equations.

I'm away from my PC but If I remember correctly the final velocity I calculated was around 6 or 7 m/s.
 
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CWatters said:
Ok but you can't calculate the impact velocity like that because the egg is accelerating not falling at a constant velocity. You need to use the equations of motion or SUVAT equations.

I'm away from my PC but If I remember correctly the final velocity I calculated was around 6 or 7 m/s.
does delta v= a*t work?
 
Yellowkies_3275 said:
does delta v= a*t work?
Yes, roughly speaking. However, it is only a rough approximation if the "a" is not constant. What "t" did you have in mind?
 
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Yellowkies_3275 said:
does delta v= a*t work?
Yes it would.

However I'm suspicious of the time you measured because it seems too quick. There is another equation that allows you to calculate the time that it should take to fall 3m knowing the acceleration.

If short of time use the value for time you measured. Otherwise perhaps look at the other equations of motion or SUVAT equation to calculate the time. Perhaps do both?
 
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