How Is the Initial Height h0 Calculated in a Spring Compression Problem?

[FallingMan]

Homework Statement

A 700 g block is released from rest at height h0 above a ver- tical spring with spring constant k = 400 N/m and negligible mass. The block sticks to the spring and momentarily stops after com- pressing the spring 19.0 cm. What is the value of h₀?

Homework Equations

W = 1/2kx^2 where W is the work, k is the spring constant, and x is the distance spring is compressed.

Potential Energy = mgh₀ where h₀ is the initial height, g is gravitational acceleration, and m is mass.

U₀ + KE₀ = U_f + KE_f , expressing that the kinetic and potential energy before must equal the kinetic and potential energy after.

The Attempt at a Solution

I have the solution for this problem, but I don't really understand it.

At height h₀ before release, the block has potential energy and no kinetic energy:
U₀ = mgh₀

At the contact point between block and spring, the block has some kinetic energy and some potential energy, which will be equal to mgh₀.

mgh₀ = KE + U

U = mgx, where x is the remaining distance to compress the spring
KE = 1/2mv^2, where v is the velocity of the block upon contact.

I know KE is normally expressed as 1/2mv^2, but the solution shows it as 1/2kx^2.

They make mgh₀ = 1/2kx^2 - mgx

This then allows them to calculate the h₀. However, I'm confused at how they arrived at that equation (specific questions below):

Why is mgx negative? Why is 1/2kx^2 expressed as the Kinetic energy? I thought 1/2kx^2 is the work done by a spring, not the kinetic energy.

 

[Panphobia]

1/2kx^2 is not kinetic energy but it is Elastic Potential Energy. Also it has no kinetic energy when upon contact, as you know kinetic energy = 1/2mv^s and when the block contacts the spring and presses it down the velocity = 0 making kinetic energy = 0, so you know that Ei = Ef, i = initial, f = final, and Ei = Ki + Ui, and Ef = Kf + Uf.

 

[haruspex]

Also it has no kinetic energy when upon contact

No, it will still have KE on contact. KE only becomes 0 when the spring is fully compressed.

They make mgh₀ = 1/2kx^2 - mgx

When the spring is fully compressed, the block has descended h₀ + x and is again stationary. So $\frac 12 k x^2 = mg(h_0+x)$

 

[Panphobia]

The block sticks to the spring and momentarily stops after com- pressing the spring 19.0 cm

That is what I meant by it contacts, I am sorry if I was misunderstood.

 

[FallingMan]

Panphobia and Haruspex,

Thank you for your clear explanation. I was looking at the block as it was contacting the spring, which was overcomplicating my problem. It seems it is much simpler to look at the problem when the kinetic energy is zero in both scenarios, namely at the very top and at the very bottom.

I guess I was also confusing 1/2kx^2 for the kinetic energy when it is the elastic potential energy.