Answer:
The pad is compressed by 5.02 cm
Explanation:
First of all, we have to calculate the initial mechanical energy of the egg, which is equal to its gravitational potential energy at the top:
[tex]E=GPE=mgh[/tex]
where
[tex]m=56.4 g = 0.0564 kg[/tex] is the mass of the egg
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
h = 14.0 m is the initial height
Substituting
[tex]E=(0.0564)(9.8)(14.0)=7.74 J[/tex]
As the egg falls down, this energy is conserved and it is entirely converted into kinetic energy; so, when the egg reaches the pad, its energy is now equal to its kinetic energy:
[tex]E=\frac{1}{2}mu^2[/tex]
where
u is the velocity of the egg as it reaches the pad
Solving for u,
[tex]u=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(7.74)}{0.0564}}=16.6 m/s[/tex]
Then, the egg is stopped by the pad, so it has a uniformly accelerated motion; so we can find the distance it takes to stop by using the suvat equation
[tex]s=(\frac{u+v}{2})t[/tex]
where:
u = 16.6 m /s is the initial velocity
v = 0 is the final velocity
[tex]t=6.05 ms = 6.05\cdot 10^{-3}s[/tex] is the stopping time
Solving,
[tex]s=(\frac{16.6+0}{2})(6.05\cdot 10^{-3})=0.0502 m = 5.02 cm[/tex]
So, the pad is compressed by 5.02 cm.
