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

AI Thread Summary
To calculate the force needed to stop an 8,000lb tractor moving at 5mph in one foot, kinetic energy (K) is determined using the formula K=1/2mv^2, resulting in 6,680 ft-lbs. The force (F) is then calculated using F=K/d, where d is 1 foot, yielding a force of 1501.7 lbs. However, the conversion to SI units was criticized for complicating the calculations. The consensus suggests focusing solely on imperial units for clarity. The discussion emphasizes the importance of accurate unit usage in physics calculations.
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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