Spring Constant to Bring a Car to Rest

AI Thread Summary
To determine the spring constant needed to bring a 1400 kg car to rest from 28 m/s with a maximum acceleration of 5g, the calculations reveal discrepancies in the approach. The initial attempt yielded a spring constant of 17150 Nm, which is significantly higher than the expected 4288 Nm. The key issue arises from the assumption of constant deceleration; the spring force increases as it compresses, meaning the acceleration is not constant throughout the stopping distance. This misunderstanding highlights the need to account for the variable nature of spring force when applying kinetic and potential energy equations. The discussion emphasizes the importance of correctly modeling the forces involved in such dynamic scenarios.
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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