Question about springs

[JFK90787]

---

Members do not see this ad.

Here's my question, I figured this out using conservation of energy rules, but my first intuition was to solve it uses F=kx since it requires less math. Anyhoo I keep getting a wrong answer using F=kx and I'm assuming I'm doing something stupid:

'A 2kg block is dropped 45cm above a spring. If the spring compresses 5cm, what is the spring constant?':

Why can I not solve this using F=20N, and x=.05m, and plugging in to F=kx ? I think the answer was 8000N/m for those who care

 

[wr8487]

Are you sure the answer was 8000N/s?

 

[wr8487]

N/m ***

 

[StarryNights]

I think you have to use conservation of energy with the elastic potential energy equation (1/2kx^2). I remember a similar pop goes the weasel problem where I got nowhere trying F=kx until I tried elastic energy instead. If you do, you'll get ~ 7800 N/m as your answer which is close enough to 8000.

Haha, nvm, saw that you got the elastic potential energy part. But yeah, my first instinct was also to do F=kx...and I kept at it for a long time but still got nowhere.

 

[JFK90787]

But yeah, my first instinct was also to do F=kx...and I kept at it for a long time but still got nowhere.

Exactly. Something is obviously wrong with the forces but I can't figure it out. It pisses me off because this is the type of thing I can totally imagine myself wasting 5 minutes on the actual test before I give up try the conservation of energy equations

 

[JFK90787]

Are you sure the answer was 8000N/s?

99% sure, I'll confirm in a few hours when I get back to my room

 

[wr8487]

This is what I did....

mgh=1/2(k)(X)^2
(2kg)(9.8m/s)(.45m)=1/2(k)(.05m)^2
K=7,840

I'm terrible at physics though so don't trust me

 

[Isoprop]

Why can I not solve this using F=20N, and x=.05m, and plugging in to F=kx ?

Let's think about what your saying. If I jump on the bed, the higher I drop onto the bed, the more the spring will be compressed right? If you just use F=kx like that, it doesn't explain why the bed spring compress MORE when I drop onto the bed springs from a higher distance.

What F=kx is saying is how much force the spring exterts when it is compressed by a distance x. We can't assume that at 5 cm, the spring will exert 20 N of force. If that was true, then the object won't bounce back up. The weight of the object will exactly equal the force from the spring, and those forces would cancel.

When you use the conservation of energy for both gravitational and the spring, you should be able to solve for k. Then use that k to see how much force is exerted at 5 cm. It should be larger than the weight of the book so that the book bounces back up.

 

[JFK90787]

What F=kx is saying is how much force the spring exterts when it is compressed by a distance x. We can't assume that at 5 cm, the spring will exert 20 N of force. If that was true, then the object won't bounce back up. The weight of the object will exactly equal the force from the spring, and those forces would cancel.

I love you long time

 

[premed101]

You can F=kx if you just placed it on the spring and wanted to solve for the spring constant.
The extra variable of dropping it makes it require the cons of energy formulas