Hooke's Law and a spring

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  • #1
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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?
 

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  • #2
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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
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I used g, yes. and I did convert.
 
  • #4
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I'm really sorry, I actually get a value of just over 3700.
 
  • #5
PPonte
Hollysmoke said:
1/2mv2^2=1/2mv1^2+mad
What formula is that?

I would do it differently. :tongue2:

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

[tex]v^2 = {v_0}^2 + 2ax[/tex]

It gives [tex]x = 8.0[/tex] m.

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

[tex]F = -k.x[/tex]

[tex]m.a = -k.x[/tex]

[tex]k = \frac{-m.a}{x}[/tex]

[tex]k = \frac{-1200 \times -5 \times 9.80}{8}[/tex]

[tex]k = 7350[/tex] N/m
 
  • #6
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neutrino said:
I'm really sorry, I actually get a value of just over 3700.
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
PPonte
Hollysmoke said:
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.
But in this case, isn't the car undergoing a constant acceleration of -5g? :grumpy:
 
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  • #8
Andrew Mason
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Hollysmoke said:
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?
You cannot use [itex]1/2mv2^2=1/2mv1^2+mad[/itex] 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: [itex]\frac{1}{2}mv^2 = \frac{1}{2}kd^2[/itex], so:

[tex]kd^2 = 5mgd = mv^2[/tex]

[tex]5gd = v^2[/tex]

[tex]d = v^2/5g = 15.75 m[/tex]

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

AM
 
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  • #9
PPonte
Ah... of course it is not constant. Sorry, my mistake.
 
  • #10
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It's a maximum acceleration
 
  • #11
nrqed
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Hollysmoke said:
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.
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
PPonte
Yes, know I understand. I was naive. :grumpy:

Thank Hollysmoke and Andrew Mason! :approve:
 
  • #13
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nrqed said:
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!
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
Andrew Mason
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Hollysmoke said:
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
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
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Okay, I'll do that. I'll tell you how it goes tomorrow ^^
 
  • #16
nrqed
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Hollysmoke said:
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
: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!
 

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