Finding a spring constant of a car

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SUMMARY

The discussion revolves around calculating the spring constant of a car with a mass of 2000 kg, which compresses the springs from a ground clearance of 40 cm to 30 cm when an additional weight is applied. The natural frequency of the springs is given as 0.6667 Hz. To find the spring constant (k), two equations are necessary due to the presence of two unknowns: the weight of the additional person and the spring constant itself. The equations involve the gravitational force and the relationship between mass, spring compression, and natural frequency.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Knowledge of natural frequency of a mass-spring system
  • Basic physics concepts related to forces and mass
  • Ability to solve simultaneous equations
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  • Learn how to derive the spring constant from mass and compression using Hooke's Law
  • Study the formula for natural frequency of a spring-mass system
  • Explore simultaneous equations to solve for multiple unknowns in physics problems
  • Investigate the effects of additional weight on spring compression in automotive applications
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Physics students, automotive engineers, and anyone interested in understanding the mechanics of spring systems and their applications in vehicles.

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Homework Statement


A car has a total mass of 2000kg, and a ground clearance of 40cm. But some fat dude got in and he made the ground clearance 30cm. What is the spring constant for this bad boy of a car? (The dude was so fat he made the shocks act like springs if you can believe it.) Also, it's a fact that the natural frequency of said springs are .6667 HZ.

Homework Equations



F = -kx

The Attempt at a Solution



So I tried being like: (2000kg)*(9.8m/s^2) = -kx
and then I was like: (2000kg + Xkg) * (9.8m/s^2) = -k(x-.1m)

I tried solving for k as hard as I could cause I really want it, but there are too many variables and I can't figure out what to do. My physics teacher says he wants me to fail and I'd like to show him that I'm not a total turd. Thank you.
 
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The car weight compresses the springs to 40cm, but it is the car+dude that compresses the springs to 30cm. So the extra weight of the dude compresses the springs an extra how much?

Your trouble is that you don't know the weight of the dude or the spring constant ... that's two unknowns: so you need two equations.

You also need the equation for the natural frequency of a mass on a spring.
 
Ok Physics bros... I think I had a stroke of genius, thanks to Simon Bridge, but I'm wondering of the mathematical/physical legality of what I've done.

So...

(2000kg)*(9.8m/s^2) = -kx = (2000kg)*(g) = -kx

(2000kg + Xkg) * (9.8m/s^2) = -k(x-.1m)= (2000kg +Xkg) * (g)

Is it legal to do this...

(-k(x-.1m)= (2000kg +Xkg) * (g)) - (-kx = (2000kg)*(g)) = (X*g = -.1k)

Then say that k = -(X*g)/.1

Cause if that's k, I can plug it into other formula and I'm good to go.
 
Are you confusing the clearance with the compression?

Let the uncompressed clearance be x0 - then the car weight Mg compresses the springs by x, giving a clearance of 40cm ... eg.

40cm = x0-x: kx=Mg ... thus: 40cm = x0-Mg/k

similarly for Car + dude: 30cm = x0-(M+m)g/k

But you don't know x0, m, or k ... so that's two equations and three unknowns.

You are still missing the equation for the natural frequency of the spring.
 

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