A 19 kg block is dragged over a rough, horizontal surface by a constant force of 174N acting at an angle of 34.9 above the horizontal. The block is displaced 54m, and the coefficient of kinetic friction is .233. The acceleration of gravity is 9.8 ms-2. Find the work done by the force of friction.
fk = [tex]\mu[/tex]n
[tex]\Sigma[/tex]Fx = Fcos[tex]\theta[/tex]-fk
The Attempt at a Solution
So I think that my online homework module is incorrect on this one because I don't see how it could be any other way. Here is what I have done:
W=[tex]\Sigma[/tex]Fx*d = [174Ncos(34.9)-(.233)(19Kg)(9.8ms-2)]*54m = 5363.37863J.
Now this is the net-work done but others online show this as the correct way to find the work done by the friction force. I tried it and it was incorrect. Then I re-read the problem and though maybe it was just fk*d which gave me 2342.7684J which is also incorrect. Where am I going wrong here or am I correct in my assumption that the online module is wrong on this one? Thanks in advance.