Mass pushed by horizontal force at constant speed on an incline

[Destrio]

3. A 52.3-kg trunk is pushed 5.95m at constant speed up a 28.0 degree incline by a constant horizontal force. The coefficient of kinetic friction between the trunk and the incline is .19 . Calculate the work done by a) the applied force and b) the force of gravity.

Fy = N - mgcos28 = 0

Fx = Fcos28 - f - mgsin28 = 0
Fx = Fcos28 - ukN - mgsin28 = 0
Fx = Fcos28 - ukmgsin28 - mgsin28 = 0
F = 282.74N

W = F*d
W = 282.74N * 5.95m = 1682.3 J

where am I going wrong?

thanks

 

[Doc Al]

Fy = N - mgcos28 = 0

The applied force will have a y-component also.

 

[Destrio]

Fy = N - Fsin28 - mgcos28 = 0
N = Fsin28 + mgcos28

Fx = Fcos28 - f - mgsin28 = 0
Fx = Fcos28 - ukN - mgsin28 = 0
N = -(mgsin28 - Fcos28)/uk
Fsin28 + mgcos28 = (-mgsin28 + Fcos28)/uk
F = -mgsin28(1-uk) / (uksin28 - cos28)
F = 245.5N

W = F*d*cos(theta)
W = 245.5N * 5.95 * cos28
W = 1289.7

I'm still getting the wrong answer, I must be making another mistake elsewhere

thanks

 

[Doc Al]

Fy = N - Fsin28 - mgcos28 = 0
N = Fsin28 + mgcos28

Fx = Fcos28 - f - mgsin28 = 0
Fx = Fcos28 - ukN - mgsin28 = 0
N = -(mgsin28 - Fcos28)/uk
Fsin28 + mgcos28 = (-mgsin28 + Fcos28)/uk

Looks OK.

F = -mgsin28(1-uk) / (uksin28 - cos28)

Check this step.

 

[Destrio]

aha!
got it

thanks very much
this problem was giving me much grief