Calculate Work Done by Forces on 330kg Piano Sliding Down 28º Incline

[tica86]

A 330-kg piano slides 3.6 m down a 28º incline and is kept
from accelerating by a man who is pushing back on it
parallel to the incline. The effective coefficient of
kinetic friction is 0.40.

a)What is the work done by the man on the piano? (in joules)

This is what I did but the answer I got is incorrect.
Wman= F*cos180 = (mgsin 28 –umgcos 28)(-1)=-376.0
I also tried F * d cos 180 = - F *3.6 = - 1353.96 J
and that is also incorrect. If anyone could help out I would
appreciate it, thanks!

b)What is the work done by the friction force? (in joules)
As for b I did the following
- 0.4 {mg cos 28} * 3.6 = - 4111.85 J
which is incorrect

 

[PhanthomJay]

A 330-kg piano slides 3.6 m down a 28º incline and is kept
from accelerating by a man who is pushing back on it
parallel to the incline. The effective coefficient of
kinetic friction is 0.40.

a)What is the work done by the man on the piano? (in joules)

This is what I did but the answer I got is incorrect.
Wman= F*cos180 = (mgsin 28 –umgcos 28)(-1)=-376.0

This is the force exerted by the man acting up the plane

I also tried F * d cos 180 = - F *3.6 = - 1353.96 J

this looks correct, but you should round off your answer to -1400 J (to 2 significant figures)

b)What is the work done by the friction force? (in joules)
As for b I did the following
- 0.4 {mg cos 28} * 3.6 = - 4111.85 J

looks good, just round it off to -4100 J

 

[tica86]

This is the force exerted by the man acting up the plane this looks correct, but you should round off your answer to -1400 J (to 2 significant figures)
looks good, just round it off to -4100 J

Thanks! It was the rounding off that got me