Question

To his surprise, Daniel found that an egg did not break when he accidentally dropped it from a height of 0.35 m onto his floor, covered with 3.0-cm-thick carpet.
Estimate the minimum acceleration of the egg while it was slowing down after touching the carpet.

Answer 1

Answer:

$-112.7 m/s^2$

Explanation:

We start by calculating the speed of the egg when it touches the upper surface of the carpet. This can be done by using the law of conservation of energy: in fact, the initial gravitational potential energy of the egg is converted into kinetic energy at the moment of impact, so we can write

$U=K\\mgh = \frac{1}{2}mv^2$

where

m is the mass of the egg

$g=9.8 m/s^2$ is the acceleration of gravity

h = 0.35 m is the initial height of the egg

v is the speed as it touches the carpet

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(0.35)}=2.6 m/s$

Now we analyze the motion of the egg through the carpet. The minimum acceleration will be obtained when the egg cover the whole tickness of the carpet to stop, that is

s = 3.0 cm = 0.03 m

And the egg must stop after covering this distance, so the final velocity will be

v' = 0

To find the acceleration, we can use the suvat equation:

$v'^2-v^2=2as$

and solving for a,

$a=\frac{v'^2-v^2}{2s}=\frac{0^2-2.6^2}{2(0.03)}=-112.7 m/s^2$