- #1

jmm5180

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## 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.