An egg in a special container together have a mass of 148.0 g.The container is designed such that when dropped it will bring the egg to a stop in 0.200 s when it hits the ground.A force of 4.91 N will break the egg.What is the maximum height from which the egg in its container can be dropped and not break the egg? (in m)
A)1.56
B)1.76
C)1.99
D)2.24
E)2.54

Question

Quizplus · Accepted Answer

First, we need to find the velocity with which the container and egg hit the ground. We can use the equation:
force*time = mass*velocity

Rearranging, we get:
velocity = force*time/mass

Plugging in the given values, we get:
velocity = 4.91 N * 0.200 s / 0.148 kg
velocity = 6.63 m/s

Now, we can use the conservation of energy principle:
potential energy at maximum height = kinetic energy just before hitting the ground
mgh = (1/2)mv^2

Solving for h, we get:
h = (1/2)v^2/g

Plugging in the given values, we get:
h = (1/2)(6.63 m/s)^2 / 9.81 m/s^2
h = 2.24 m

Therefore, the maximum height from which the egg in its container can be dropped without breaking the egg is 2.24 m, which is answer choice D.

An Egg in a Special Container Together Have a Mass