Spring Question -- A mass falls onto a spring compressing it....

[smartdude00111]

Homework Statement

An object with a mass of 6kg falls 1.8m onto a spring with a spring constant of 2x10^3 N/m. At that instant, the velocity of the object is 4m/s. What is the compression of the spring and the max displacements of the spring.
m = 6kg
v = 4m/s
h = 1.8m
k = 2x10^3N/m

Homework Equations

Gravitational potential = mgh
Elastic potential = (1/2)kx^2
Kinetic energy = (1/2)mv^2[/B]

The Attempt at a Solution

(1/2)kx^2 = (1/2)mv^2
x = √((mv^2)/k)
x = √(((6)(4)^2)/2x10^3)
x = 0.219
This is my answer to a test question but many other people did it in different ways. Some used the floor as the reference point for y=0 where as I used the top of the spring at equilibrium.

 

[Doc Al]

The gravitational PE changes as the mass falls. You cannot ignore it.

 

[smartdude00111]

The gravitational PE changes as the mass falls. You cannot ignore it.

We’re using the law school of conservation of energy. So if the reference point is at the top of the spring, the gravitational potential is 0, so the only energy there is the kinetic which is equal to the elastic potential when the spring is fully compressed

 

[Doc Al]

So if the reference point is at the top of the spring, the gravitational potential is 0, so the only energy there is the kinetic which is equal to the elastic potential when the spring is fully compressed

If the gravitational PE at the top is 0 (a perfectly OK reference point), what will it be when the spring is compressed? It changes.