Question

If you have a 60 g egg dropped from a height of 10 m with cushioning material that delays the impact by 2 seconds what was the force exerted on the egg by the floor?

Answer 1

The potential energy of the egg when it begins to fall is equal to the kinetic energy of the egg just before touching the ground:
m * g * h = (1/2) * m * v ^ 2
Clearing the speed:
v = root (2 * h * g)
Substituting the values
v = root (2 * (10) * (9.8)) = 14
Then, by definition, we have
F * deltat = m * deltav
Clearing F
F = (m * deltav) / (deltat)
Substituting the values
F = ((0.060) * (14)) / (2) = 0.42
answer
the force exerted on the egg by the floor was 0.42N