# Hooke's Law and a spring

1. Jul 26, 2006

### Hollysmoke

Okay, I'm having a debate with my teacher. He's saying I'm wrong but I still think I'm right. The question is:

What should be the spring constant, k, of a spring designed to bring a 1200kg car to rest from a speed of 100km/h so that the occupants undergo a maximum acceleration of 5.0g?

He said to use the equations v2^2 = v1^2 + 2ad and 1/2mv2^2=1/2mv1^2+mad

I said ma=kx, ma/k=x

then 1/2mv^2=1/2k(ma/k)^2, solve for k and I get 3700, whereas he got 14,000. Could someone please help me out with this?

2. Jul 26, 2006

### neutrino

Infact, I get a value of just over 14,000 using your method. Did you convert the speed to m/s? Also use g = 9.8 ms-2.

3. Jul 26, 2006

### Hollysmoke

I used g, yes. and I did convert.

4. Jul 26, 2006

### neutrino

I'm really sorry, I actually get a value of just over 3700.

5. Jul 26, 2006

### PPonte

What formula is that?

I would do it differently. :tongue2:

First, I used this formula do find the displacement of the spring:

$$v^2 = {v_0}^2 + 2ax$$

It gives $$x = 8.0$$ m.

Then, I used Hooke's Law to find the spring constant.

$$F = -k.x$$

$$m.a = -k.x$$

$$k = \frac{-m.a}{x}$$

$$k = \frac{-1200 \times -5 \times 9.80}{8}$$

$$k = 7350$$ N/m

6. Jul 26, 2006

### Hollysmoke

Yes, I did too!

See, my reasoning is, F=ma=kx, but k and m are constant, therefore when a greater force is applied to scretch the spring (or in this case, compress), acceleration changes (as well as x). However, the other two equations involves a uniform acceleration, which you can't have in this case.

7. Jul 26, 2006

### PPonte

But in this case, isn't the car undergoing a constant acceleration of -5g? :grumpy:

Last edited by a moderator: Jul 26, 2006
8. Jul 26, 2006

### Andrew Mason

You cannot use $1/2mv2^2=1/2mv1^2+mad$ because the acceleration is not constant.

If the maximum acceleration is 5 g = -49 m/sec^2 then kd = 5mg where d is the stopping distance. So k = 5mg/d

The work done in stopping the car is stored in spring: $\frac{1}{2}mv^2 = \frac{1}{2}kd^2$, so:

$$kd^2 = 5mgd = mv^2$$

$$5gd = v^2$$

$$d = v^2/5g = 15.75 m$$

So k = 5mg/d = 5*1200*9.8/15.75 = 3700 N/m to two significant figures.

AM

Last edited: Jul 26, 2006
9. Jul 26, 2006

### PPonte

Ah... of course it is not constant. Sorry, my mistake.

10. Jul 26, 2006

### Hollysmoke

It's a maximum acceleration

11. Jul 26, 2006

### nrqed

I can't check the numbers right now but your method seems correct. And of course, as you pointed out, the equation he suggested is wrong because the acceleration is not constant. I am sure that when you tell him/her that, he will slap himself and agree with you!

12. Jul 26, 2006

### PPonte

Yes, know I understand. I was naive. :grumpy:

Thank Hollysmoke and Andrew Mason!

13. Jul 26, 2006

### Hollysmoke

Actually, that's why I'm here. I told him and he brushed me off, saying F=ma=kx <- Acceleration is constant. I actually just stormed out of the class and had a discussion with the VP about it. I just wanted to make sure I wasn't making a total ass of myself though. Thanks for your help :3

14. Jul 26, 2006

### Andrew Mason

Ask him how the force can be constant if x changes. If he says it doesn't change, ask him to explain physically how the spring slows the car down. Perhaps there we have a different understanding of how the spring is configured.

AM

15. Jul 26, 2006

### Hollysmoke

Okay, I'll do that. I'll tell you how it goes tomorrow ^^

16. Jul 26, 2006

### nrqed

:surprised :surprised :surprised :surprised
I am flabbergasted!!
I mean, I can imagine a prof being distracted and for a second using an invalid equation, but to *defend* that the acceleration is constant for mass connected to a spring is amazing. Wow...

Good luck!