Can you confirm this answer?

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I don't have an answer to this, as I'm going over an old test, but I just want to make sure I'm doing it right.

A 120kg crate is pulled from rest across a frictionless floor with a constant horizontal force of 400 N for a distance of 12m. The force continues to be applied, but for the next 12 m the floor is not frictionless and has a coefficient of kinetic friction of .30. What's the final speed?

I get 9.44m/s by adding the work done when the box is frictionless (4800J), then subtracting it from the work lost to the friction (4233) and set that quantity = .5mv^2. Is this right?
 

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  • #2
dextercioby
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psycovic23 said:
I don't have an answer to this, as I'm going over an old test, but I just want to make sure I'm doing it right.
A 120kg crate is pulled from rest across a frictionless floor with a constant horizontal force of 400 N for a distance of 12m. The force continues to be applied, but for the next 12 m the floor is not frictionless and has a coefficient of kinetic friction of .30. What's the final speed?
I get 9.44m/s by adding the work done when the box is frictionless (4800J), then subtracting it from the work lost to the friction (4233) and set that quantity = .5mv^2. Is this right?
Nope.U need to consider this statement:(underlined).U'll figure out what's missing from your judgements.

Daniel.
 
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Hm....I still can't get it. After it's in the friction zone, there's 400N forward * 12m to give you 4800J - 352.8N of friction * 12m = the work done on the box during the duration of second half of sliding. You add that to the first half...solve for .5mv^2 and I get 9.45 again...I'm lost :cry:
 
  • #4
dextercioby
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psycovic23 said:
Hm....I still can't get it. After it's in the friction zone, there's 400N forward * 12m to give you 4800J - 352.8N of friction * 12m = the work done on the box during the duration of second half of sliding. You add that to the first half...solve for .5mv^2 and I get 9.45 again...I'm lost :cry:
If u wrote anything similar to this:
[tex] \frac{mv^{2}}{2} =9600 -\mu mg\cdot 12 [/tex]
,then it's okay.

Daniel.

PS.I believe that 9,45m/s is okay.
 
  • #5
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Yea, that's what I had..although in a slightly different form. Alright, I'm happy to know I wasn't wrong now :biggrin: I was rippin my hair out over that! :tongue2:
 

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