Ricky Bobby: High Speed Adventure on the Highway

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Ricky Bobby??

Homework Statement



Ricky Bobby jumps into his 1003kg pick up and heads out at 69.2mi/h. It's night and he notices too late the warning sign that the bridge is out. He slams on the brakes and skids for 45.2(kinetic friction=0.78)before hitting the long massive coil spring that is rigidly mounted onto the barricade. In addition to his skidding tires, the spring brings him to a stop. The spring constant is 3.57 x 10^4 N/m. how far does he travel after making contact with spring?

Homework Equations


1/2mv^2+1/2Kx^2=1/2mv^2+1/2kx^2 ?// I am not sure if this is the right equation i don't think it is I've looked everywhere in my book to find a relevant equation i was wondering if anyone can point me towards the right equation please


The Attempt at a Solution



refer to (b)2.
 
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ldbaseball16 said:
Ricky Bobby jumps into his 1003kg pick up and heads out at 69.2mi/h. It's night and he notices too late the warning sign that the bridge is out. He slams on the brakes and skids for 45.2(kinetic friction=0.78)before hitting the long massive coil spring that is rigidly mounted onto the barricade. In addition to his skidding tires, the spring brings him to a stop. The spring constant is 3.57 x 10^4 N/m. how far does he travel after making contact with spring?

1/2mv^2+1/2Kx^2=1/2mv^2+1/2kx^2 ?// I am not sure if this is the right equation

Hi ldbaseball16! :smile:

Is that 45.2 feet?

This is an energy problem …

so yes, that equation is right, but you also need to use the work-energy theorem … work done by friction = loss of mechanical energy :wink:
 


ok Wnet=(1/2)mv^2-(1/2)mv^2 this is the work energy theorem but on the other equation i have to apply the kinetic friction to 45.2m? right and once i find the conservation of mechanical energy what do i do with that number? how do i apply it to the work-energy theorem?
 
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ldbaseball16 said:
ok Wnet=(1/2)mv^2-(1/2)mv^2 this is the work energy theorem but on the other equation i have to apply the kinetic friction to 45.2m? right

uhh? :confused: too cryptic :redface:

just do it! :rolleyes:
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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