Spring constant k, car to rest

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SUMMARY

The spring constant k required to bring a 1440 kg car to rest from a speed of 111 km/h, ensuring occupants experience a maximum acceleration of 5.0 g, is calculated to be approximately 7274.23 N/m. The calculations involve using the equations F = -kx, F = ma, and the kinematic equation v² = v(initial)² + 2a(x). The maximum acceleration occurs at maximum deflection, necessitating the use of average acceleration to determine the spring's deflection and energy conservation principles to finalize the calculations.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Familiarity with Hooke's Law (F = -kx)
  • Knowledge of kinematic equations
  • Basic principles of energy conservation in mechanics
NEXT STEPS
  • Explore the derivation of Hooke's Law and its applications in real-world scenarios.
  • Study the relationship between kinetic energy and potential energy in spring systems.
  • Learn about the effects of varying acceleration on spring dynamics.
  • Investigate advanced topics in dynamics, such as damping and oscillations in spring systems.
USEFUL FOR

Physics students, mechanical engineers, and anyone involved in automotive safety design or spring mechanics will benefit from this discussion.

len2
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Homework Statement


What should be the spring constant k of a spring designed to bring a 1440 kg car to rest from a speed of 111 km/h so that the occupants undergo a maximum acceleration of 5.0 g?
____N/m

Homework Equations


F=-kx
F=ma
V2=v(initial)2 + 2a(x)
v=v(initial) + at


The Attempt at a Solution


a=5 x 9.8 = 49
F= 1440 x 49 = 70560
x= 30.83^2 +2(-49)x = 9.7
F/x= k =70560/9.7 = 7274.23
 
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Your reasoning looks correct. Is there a problem you were having or did you just need a check on your method?
 
len2 said:

Homework Statement


What should be the spring constant k of a spring designed to bring a 1440 kg car to rest from a speed of 111 km/h so that the occupants undergo a maximum acceleration of 5.0 g?
____N/m

Homework Equations


F=-kx
F=ma
V2=v(initial)2 + 2a(x)
v=v(initial) + at


The Attempt at a Solution


a=5 x 9.8 = 49
F= 1440 x 49 = 70560
x= 30.83^2 +2(-49)x = 9.7
F/x= k =70560/9.7 = 7274.23

This is mostly correct, however the equation vfinal2 = vinitial2 + 2ax is based upon the acceleration (or deceleration) a being constant over distance x. In the case of a spring the force F is proportional to the deflection, F = kx, so the F and the resulting acceleration will vary with distance,

F = ma = kx or a = (k/m)x.

The maximum acceleration occurs at maximum deflection, and in this problem the maximum or limiting acceleration is 5g (or ~ 49 m/s2).

There are two parameters which one must find in this problem, one is the spring constant and the other is the maximum deflection which is not given. One needs two 'independent' equations to solve for two parameters.

Now to finish the problem, find the 'average' acceleration which gives 0 for x=0, and 5g for x = d, where d is the maximum or total displacement of the spring. So, if a(x) = 0, and a(d) = 5g, and a is linear in x, then the average a is just the arithmetic average <a> = 5g/2, or the average a = <a> = amax/2. Then one can use

vfinal2 = vinitial2 + 2ad, where d is the distance over which the average a is applied.

or realizing the previous equation is a form of the conservation of energy, one can realize that the initial kinetic energy of the car must be transformed into the mechanical potential energy stored in the spring at full deflection, i.e.

1/2 mv2 = 1/2 k d2, or

mv2 = k d2.

Using Fmax = m amax = k d, find d and substitute it into the other equation.

Here is a handy reference - http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html

http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html
 

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