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

[psycovic23]

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?

 

[dextercioby]

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.

 

[psycovic23]

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

 

[dextercioby]

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

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.

 

[psycovic23]

Yea, that's what I had..although in a slightly different form. Alright, I'm happy to know I wasn't wrong now I was rippin my hair out over that! :tongue2: