Spring constant k, car to rest

In summary, the spring constant for a car to come to rest from a speed of 111 km/h and experience a maximum acceleration of 5g is 7274.23 N/m.
  • #1
len2
1
0

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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  • #2
Your reasoning looks correct. Is there a problem you were having or did you just need a check on your method?
 
  • #3
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
 

Related to Spring constant k, car to rest

What is the spring constant, k?

The spring constant, k, is a measure of the stiffness of a spring. It is defined as the force required to stretch or compress a spring by a certain distance.

How is the spring constant, k, calculated?

The spring constant, k, can be calculated by dividing the force applied to the spring by the distance the spring is stretched or compressed. Mathematically, it is expressed as k = F/x, where F is the applied force and x is the displacement of the spring.

What is the unit for spring constant, k?

The unit for spring constant, k, is Newtons per meter (N/m) in the SI system of units. In other systems of units, it may be expressed as pounds per inch (lb/in) or dynes per centimeter (dyn/cm).

How does the spring constant, k, affect the motion of a car coming to rest?

The spring constant, k, affects the motion of a car coming to rest by determining the amount of force required to stop the car. A higher spring constant means a stiffer spring, which requires more force to compress and therefore slows down the car more quickly.

Can the spring constant, k, change?

Yes, the spring constant, k, can change depending on factors such as the material and shape of the spring. It can also change if the spring undergoes permanent deformation due to overstretching or compression.

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