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

[Eric Ahlvin]

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

 

[Eric Ahlvin]

Sorry, I see this was not the place to put this, but I also do not see how to delete it...

 

[Bystander]

It makes a great homework problem. Leave it right where it is and don't worry.

K= 6,680 ftlbs

This is good.

where F=K/d and d=1ft

This is good.

F= 6680Newtons and 1 Newton = 0.224808943 pounds force

What are you trying to do dragging in SI units? NOT good.

so F= 1501.7 lbs

The SI conversion messed this up for you. Try it again without worrying about SI.