1. The problem statement, all variables and given/known data A 10 kg block lies on an incline 15 degrees above the horizontal. A force F, parallel to the incline, pushes the block up the incline. There is a displacement of 7m and the displacement is done with constant velocity. Given that the coefficient of kinetic friction, μ, is 0.2, determine the magnitude of the force F and determine the work done on the block by all the forces in the system. 3. The attempt at a solution http://i.imgur.com/N4nQMH4.png The object's in equilibrium, so [tex] \sum F_y = 0 [/tex] [tex] n = wcos(θ) = mgcos(θ) [/tex] θ = 15, m = 10 kg, and g = 9.8 m/s^2, so n is around 94.66 N [tex] \sum F_x = 0 [/tex] [tex] F - f - wsin(θ) [/tex] [tex] F = μn + mgsin(θ) [/tex] n = 94.66N, μ = 0.2, m = 10 kg, g = 9.8 m/s^2, θ = 15, so F is around 44.29N and f is around 18.93 N. The work done by the normal force is zero since it is perpendicular to the displacement. The work done by the force F is the product of its magnitude and the displacement since it lies parallel to the displacement. [tex] W_F = (44.29N)(7m) = 310 J [/tex] Friction works in opposition to the displacement, so it does negative work on the object. [tex] W_f = (18.93N)(7m) = - 132 J [/tex] The work done on the object by the weight is the magnitude of the component in the direction of the displacement. The weight's component in the direction of the displacement opposes the displacement, so it is also negative. [tex] W_w = (mgsinθ)(7m) = (25.36N)(7m) = - 177 J [/tex] This was a problem I had on a test and it seemed very simple and straightforward. I was convinced I didn't do anything incorrectly, but when I turned it in I saw that my answers were different than my prof's. I've looked it over again and I'm almost convinced I didn't do anything wrong. Could anyone possibly look it over and tell me if there's anything wrong with my reasoning?