1. Dec 9, 2007

### cmarte01

1. The problem statement, all variables and given/known data

A father pushes horizontally on his daughter's sled to move it up a snowy incline, as illustrated in the figure, with h = 3.9 m and = 11°. The total mass of the sled and the girl is 35 kg and the coefficient of kinetic friction between the sled runners and the snow is 0.20. If the sled moves up the hill with a constant velocity, how much work is done by the father in moving it from the bottom to the top of the hill?

m=35 kg
h=3.9 m
angle= 11 degrees
kf= .20

2. Relevant equations

F=Uk
W=Fd
W=mg
sin=OPP/HYP

3. The attempt at a solution

W= mg, W=(35)(9.8) = 343
Fx= 343 sin 11 = 65.45
Fy= 336.70
.2*336.7 = 67.34
67.34+65.45 = 132.79
sin 11 = 3.9/HYP, HYP = 3.9/sin 11 = 20.44
132.79* 20.44 = 2714.23 J

What did I do wrong???? Help!!

2. Dec 9, 2007

### Shooting Star

I can't see any figure. I assume that the vertical height of the hill is h = 3.9 m.

Component of mg acting along the slope is mgsin x, where x= 11 deg.
Normal reaction is mgcos x.
So, frictional force is kmgcos x along the slope.

Total force along the slope is the sum of these two forces, both acting downward.

W = f*d, where h/d = sin x.

Plug in the values now.