Calculating Final Speed of a 120kg Crate on a Frictionless and Frictional Floor

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A 120kg crate is pulled across a frictionless floor with a constant force of 400 N for 12m, resulting in 4800J of work. When the crate moves onto a frictional surface with a coefficient of kinetic friction of 0.30, the frictional force is calculated to be 352.8N, leading to a work loss of 4233J over the same distance. The final speed is calculated by combining the work done on the crate and the work lost to friction, resulting in a speed of approximately 9.45m/s. The calculations confirm that the approach is correct, and the final speed is validated by multiple participants in the discussion. The consensus is that the calculated speed is accurate and aligns with the physics principles involved.
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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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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.
 
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:
 
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:
\frac{mv^{2}}{2} =9600 -\mu mg\cdot 12
,then it's okay.

Daniel.

PS.I believe that 9,45m/s is okay.
 
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! :-p
 

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