Spring Constant to Bring a Car to Rest

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Homework Help Overview

The discussion revolves around determining the spring constant \( k \) required to bring a 1400 kg car to rest from a speed of 28 m/s, ensuring that the occupants experience a maximum acceleration of 5g. Participants are exploring the relationship between kinetic energy, potential energy in the spring, and the implications of deceleration on the spring's behavior.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the equations of motion and energy conservation principles, questioning the validity of their setups and calculations. They express confusion over the large value of the spring constant derived from their calculations and consider the implications of constant versus variable deceleration.

Discussion Status

There is ongoing exploration of the problem, with participants questioning the assumptions made regarding constant acceleration and the nature of the forces involved. Some guidance has been offered regarding the relationship between force, acceleration, and the spring's behavior, but no consensus has been reached.

Contextual Notes

Participants note that the problem involves a spring mechanism, which inherently leads to a variable force as the spring compresses, contrasting with the constant acceleration assumed in basic kinematic equations. This aspect is under scrutiny as they attempt to reconcile their findings with the expected outcomes.

kieth89
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Homework Statement


What should be the spring constant k of a spring designed to bring a 1400 kg car to rest from a speed of 28 m/sec so that the occupants undergo a maximum acceleration of 5g’s?

I have the solution manual and can see how they did this, but am curious as to why my attempt did not work. It didn't seem to break any rules that I can see.

a = -5g = -49 {\frac{m}{s^{2}}} neg. due to deceleration
m = 1400 kg, v_{0} = 28 {\frac{m}{s}}, v = 0 {\frac{m}{s}}

Homework Equations


I went about the problem in a roundabout way. The equations used are:
v = v_{0} + at
x - x_{0} = v_{0}t + {\frac{1}{2}}at^{2}
KE = PE
{\frac{1}{2}}mv^{2} = {\frac{1}{2}}Kx^{2}

The Attempt at a Solution


v = v_{0} + at
0 = 28 + -49t
t =0.57143 s
x - x_{0} = v_{0}t + {\frac{1}{2}}at^{2}
x - 0 = (28)(0.57143) + {\frac{1}{2}}(-49)(0.57143)^{2}
x = 8
{\frac{1}{2}}(1400)(28)^{2} = {\frac{1}{2}}K(8)^{2}
K=17150 Nm

I believe the correct answer is 4288 Nm based on a similar problem in the solution manual. I don't understand why working it like this does not work though. All the units work out, and while I know that's not a guarantee of correct setup it is usually a sign you are moving in the right direction. Anything stand out in particular here?

Thanks!
 
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kieth89 said:

Homework Statement


What should be the spring constant k of a spring designed to bring a 1400 kg car to rest from a speed of 28 m/sec so that the occupants undergo a maximum acceleration of 5g’s?

I have the solution manual and can see how they did this, but am curious as to why my attempt did not work. It didn't seem to break any rules that I can see.

a = -5g = -49 {\frac{m}{s^{2}}} neg. due to deceleration
m = 1400 kg, v_{0} = 28 {\frac{m}{s}}, v = 0 {\frac{m}{s}}

Homework Equations


I went about the problem in a roundabout way. The equations used are:
v = v_{0} + at
x - x_{0} = v_{0}t + {\frac{1}{2}}at^{2}
KE = PE
{\frac{1}{2}}mv^{2} = {\frac{1}{2}}Kx^{2}

The Attempt at a Solution


v = v_{0} + at
0 = 28 + -49t
t =0.57143 s
x - x_{0} = v_{0}t + {\frac{1}{2}}at^{2}
x - 0 = (28)(0.57143) + {\frac{1}{2}}(-49)(0.57143)^{2}
x = 8
{\frac{1}{2}}(1400)(28)^{2} = {\frac{1}{2}}K(8)^{2}
K=17150 Nm

I believe the correct answer is 4288 Nm based on a similar problem in the solution manual. I don't understand why working it like this does not work though. All the units work out, and while I know that's not a guarantee of correct setup it is usually a sign you are moving in the right direction. Anything stand out in particular here?

Thanks!
Your answer is 4 times the correct answer.

With your answer, what force does the spring exert at maximum compression ?
 
Since the car travels 8 meters to stop given a constant deceleration of 5g wouldn't the max compression be 8m? Which would make the F=17150*8, a very large number..

But we took time into account when we found the x distance, so I'm not understanding why the k constant is so large. It feels like it's giving me what k would be if it needed to decelerate the car at 1 specific moment instead of over a span of time... But I know that last kinetic energy equation is right as they are both conservative forces.
 
kieth89 said:
Since the car travels 8 meters to stop given a constant deceleration of 5g wouldn't the max compression be 8m? Which would make the F=17150*8, a very large number..

But we took time into account when we found the x distance, so I'm not understanding why the k constant is so large. It feels like it's giving me what k would be if it needed to decelerate the car at 1 specific moment instead of over a span of time... But I know that last kinetic energy equation is right as they are both conservative forces.
Take that force and divide it by the mass to find the acceleration, as the car comes to rest.
 
Oh. So the deceleration cannot be constant if it is only a spring causing it then due to the increasing force as it compresses more? Is that why the basic kinetic equations fail here, because they're finding a constant acceleration?
 
kieth89 said:
Oh. So the deceleration cannot be constant if it is only a spring causing it then due to the increasing force as it compresses more? Is that why the basic kinetic equations fail here, because they're finding a constant acceleration?
Who is "they" ?

I believe that your solution does make the average acceleration equal to -5g.
 

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