The Speed of a Block (Springs)

1. Nov 3, 2005

Ike

A 0.164-kg block is dropped straight downwards onto a vertical spring. The spring constant of the spring is 62 N/m. The block sticks to the spring and the spring compresses 0.17 m before coming to a momentary halt. What is the speed of the block just before it hits the spring? Give your answer in (m/s).

My initial thought was to use a conservation of energy formula. I came up with:

(1/2)(k)(x)^2 = (1/2)(m)(v)^2
(1/2)(62 N/m)(0.17 m)^2 = (1/2)(0.164 kg)(v)^2
v = {[(62 N/m)(0.17 m)^2]/(0.164 kg)}^(1/2)​

This, however, is not the correct answer. Therefore, there must be another way of doing this problem. Can anyone help me?

2. Nov 3, 2005

Fermat

Did you add on the loss of PE in the block due to the distance travelled - i.e. the compression of the spring ?

3. Nov 3, 2005

Ike

Well, I was assuming that at the moment just before the block hits the spring, the total energy is going to be kinetic. This is because the spring has not yet been compressed. Once the spring has been compressed as far as it will go (velocity = 0), all of the energy will be in spring potential.

Therefore, the kinetic energy before the block hits the spring will equal the potential energy of the spring at the bottom of its motion.

Is this correct?

4. Nov 3, 2005

Fermat

Just before the block hits the spring, you can consider the total energy of the block, at that point, as kinetic energy. All of this energy is absorbed by the spring during compression.

But consider the block simply placed on top of the spring, motionless, then released. The spring will compress, by an amount x say. The work done on the spring is (1/2)kx² and comes from the loss in PE of the block. That PE is mgx.

So, in your case, not only will KE from the block's movement give compression, so also will the loss in PE of the block.
The PE in the spring will then be equal to the loss in KE of the block plus the loss in PE of the block.

5. Nov 3, 2005

Ike

Okay, I see what you mean now... So I need to find how far the spring is compressed by the block if it were at rest on the spring. Then I need to subtract the resultant energy loss from the total energy of the system. Thanks for your help!

6. Nov 3, 2005

Fermat

Not quite!

Sorry, I may have been confusing you.

That was just meant as an illustration of how the loss of PE in the block, as the spring compresses, should be taken into account.

Let the spring compress by an amount x.

WD on spring = ½kx²
Loss of PE = mgx
Loss of KE = ½mv²

Energy Balance
============
½kx² = mgx + ½mv²

k = 62 N/m
x = 0.17 m
m = 0.164 kg

solve for v.