Object brought to a stop in a distance of 1 mm?

[Sarah0001]

Homework Statement

An egg is dropped at rest from a tree
c) an egg strikes the ground and is brought to a stop in a distance
of 1 mm. Assuming a mass of 20 g for the egg calculate the force
required. (You may assume a constant braking force).

I am confused on the visual picture of what is happening, when the is 'brought to a stop in a distance of 1 mm' from the moment it strikes the ground.

Relevant Equations

Know velocity of egg when it reaches the ground is 19.79 m/s Know
can use suvat or energy conservation 1/2mv^2=Fd

s=1 *10^-3 v^2 = u^2 + 2as a = v^2 - u^2 / 2*s = - 195 822.05 m/s^2
u=19.79 m/s F= 20*10^-3 *195 822.05 = 4000 N (1 s.f)
v=0
a=?or 1/2mv^2=Fd ----> F= mv^2/2d = 4000 (1 s.f)With either method, I am still confused on the visual picture of what is happening, when the egg is 'brought to a stop in a distance of 1 mm'. I have a few ideas: Does this mean that the egg bounces from the ground with initial velocity 19.79 m/s to zero at 1 mm off the ground? Or does it mean the egg possibly deforms upon hitting the ground? Or the ground deforms very slightly so the egg moves a distance of 1mm upon striking it? I'm quite confused at what I'm supposed to picture here.

 

[ZapperZ]

You know the initial velocity of the egg when it hits the ground. You know that it comes to a stop, so you know the final velocity. You know that this all happens over a distance of 1mm. You should be able to use one of the kinematical equations to find the acceleration (hint: it's the one with the square of the velocities).

Once you know acceleration a, you can find the force exerted on the egg since you know its mass.

Zz.

 

[Sarah0001]

You know the initial velocity of the egg when it hits the ground. You know that it comes to a stop, so you know the final velocity. You know that this all happens over a distance of 1mm. You should be able to use one of the kinematical equations to find the acceleration (hint: it's the one with the square of the velocities).

Once you know acceleration a, you can find the force exerted on the egg since you know its mass.

Zz.

yes..That's exactly what I did. My question is what is happening, when the egg is 'brought to a stop in a distance of 1 mm' when it hits the ground. What is the motion like then? Does it rebound, does the ground deform very slightly?

 

[haruspex]

does the ground deform very slightly?

Yes. In practice, there would be a bit of bounce, but that is after the egg has come (instantaneously) to rest, so is irrelevant. The egg may also deform a little. Which deforms the more depends on the ground surface.

 

[neilparker62]

The egg may also deform a little.

Putting it mildy I would say

An alternate approach would be to calculate the time from $v_{av}Δt=0.001$. Then $F=\frac{Δp}{Δt}$

 

[FactChecker]

I wouldn't overthink it just because of this problem. The problem has been greatly simplified so that you can just plug in some numbers and get an answer using the equations of the class. The reality of what would happen to the egg is very complicated. It depends on the properties of the shell, the egg yolk, and the rest of the egg.