What is the force needed to stop an 8,000lb tractor moving at 5mph in one foot?

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SUMMARY

The discussion focuses on calculating the force required to stop an 8,000 lb tractor moving at 5 mph over a distance of one foot. The kinetic energy formula, K = 1/2 mv², was applied, resulting in a kinetic energy of 6,680 ft-lbs. The force was then calculated using F = K/d, yielding a force of 1,501.7 lbs. However, the introduction of SI units caused confusion, indicating that calculations should remain in imperial units for clarity.

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Eric Ahlvin
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I have a situation where I need to figure how much force it will take to stop an 8,000lb tractormoving at a max speed of 5mph in somewhere around a foot. It will not be an impact stop as there is nothing for it to hit but rather a heavy brake stop. It has been quite some time since I have dealt with kinetic energy equations. Can someone check my work on this?

K=1/2mv^2
m=8,000lb x 0.0310809502 lbs/slug = 248.64slugs
v = 5m/hr x 5280ft/mile x 1hr/60min x 1min/60sec = 7.33 ft/sec

K=0.5 * 248.64slugs * 7.33^2

K= 6,680 ftlbs

where F=K/d and d=1ft

F= 6680Newtons and 1 Newton = 0.224808943 pounds force

so F= 1501.7 lbs

If someone could verify this that would be very much appreciated
 
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Sorry, I see this was not the place to put this, but I also do not see how to delete it...
 
It makes a great homework problem. Leave it right where it is and don't worry.
Eric Ahlvin said:
K= 6,680 ftlbs
This is good.
Eric Ahlvin said:
where F=K/d and d=1ft
This is good.
Eric Ahlvin said:
F= 6680Newtons and 1 Newton = 0.224808943 pounds force
What are you trying to do dragging in SI units? NOT good.
Eric Ahlvin said:
so F= 1501.7 lbs
The SI conversion messed this up for you. Try it again without worrying about SI.
 

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