# Can somebody check my work see if I did the steps right?

1. Oct 13, 2009

### Jac8897

A boy pulls a 14.6kg sled 12.5m at a constant speed across a horizontal floor. if the rope makes a 32.4degrees?
angle with the horizontal and the coefficient of friction is .489, what is the work done by each of the forces acting on the sled?
2 days ago - 2 days left to answer.
my work,
from the free body diagram I get

EFy= n+Fsin@-mg=0 "@ = theta"
EFx= Fcos@-f=0 "constant speed"

solve Fy= n=mg-Fsin@
plug it into Fx

friction=un
Fcos@-u(mg-Fsin@)=0
Fcos@-umg+uFsin@=0 >>
Fcos@+uFsin@=umg >>
F(cos@+usin@)=umg>>
F=(umg)/(cos@+usin@)
F=63.3N >>

work of the force
WF=63.3N(12.5)cos0>> WF= 791J

no solve by friction
Fcos@-f=0 >>> f=Fcos@
f=(63.3N)cos(32.5) >> f=53.4N

work of friction
Wf=(53.4N)(12.5M)cos180 >> Wf= -668J

I ignore the N force and the gravety and Fsin@ because they are perpendicular.

is my physic process good?
is my answer right?
thanks

2. Oct 13, 2009

### willem2

All the forces you calculated are OK. The total work done on the sled isn't 0. This isn't possible since the sled gains no kinetic or potential energy. Are you sure that both of the forces involved are pulling in the same direction as the motion of the sled?

3. Oct 13, 2009

### Jac8897

so the total work is W=WF+Wf

W=791J-668J
W=123J
the total work done on the sled is 123J