How much work done by friction when a box w/apparent mass...?

[jmm5180]

Homework Statement

How much work is done by friction when a box with an apparent mass of 325kg moves horizontally across a floor with coefficient of friction of .33 and is pulled at 59.9 degree angle for 2.60 m? What is the force that is pulling the box?

Homework Equations

W=Fdcos@ (@= theta)
Fnet=ma

The Attempt at a Solution

I searched for what apparent mass was, but what I found seemed to do with volume and water displacement, which aren't related to this problem. Instead I tried solving the second question first because I needed Normal force to solve the work done by friction.

For my free body diagram, I had force applied (Fa) in the y direction upwards on the y-axis along with normal force. Along the negative y-axis I had weight. Force of friction (Ff) was along the negative x axis, and force applied in the x direction was on the positive x axis.

In the x direction...
Ff=Facos@ (since no acceleration)
uN = FaCos@

In the y direction...
Fasin@ + N = W (again since no acceleration)
N= mg - FaSin@

So plugging in,,,
u(mg-Fasin@) = FaCos@

Solving for Fa I got...
umg-uFasin@ = Facos@
Fa= umg/(usin@ + cos@)
Fa= (.33*325 kg*9.8)/(.33sin59.9+cos59.9)
Getting that the force applied = 1335.50 N, but this is not correct.

Any help would be appreciated, thanks in advance.

 

[TomHart]

I found this on a Google search:
In physics, apparent weight is a property of objects that corresponds to how heavy an object is. The apparent weight of an object will differ from the weight of an object whenever the force of gravity acting on the object is not balanced by an equal but opposite normal force.

 

[anlon]

If you assume $apparent \ weight = mg - N$ do you see a way to proceed?

 

[jmm5180]

If you assume $apparent \ weight = mg - N$ do you see a way to proceed?

Yes, thank you!

 

[Bharat999]

Try taking the apparent weight = N.