My friend and I are working on a worksheet together and she totally lost me and I would like to know how she got her info.. Can you help sifted through her stuff? Here is the question: A 2.kg block on a frictionless plane inclined at 30 degrees from horizontal has a force F acting on it upward parallel to the incline. The block is accerlating downward at 2.0 m/s. Find F. Now here is my work: I assumed it was the F=ma eqation F = 2kg*2.0 = 4N I know its not that easy and cosine of 30 degress comes in somewhere.. Here is her work she sent me in an email: Now you know the force down the inclined plane. On top of that, you know that there is a force acting up the inclined plane of F. You know that: (2kg)*g*cos(60 degrees) - F = (2kg)*(2 m/s/s) That is, assuming that "downward" is the positive direction, then (2kg)*g*cos(60 degrees) is a positive force and F is a negative force. If you sum the forces, you get the net force. That net force is what is causing the 2 m/s/s acceleration. The magnitude of the net force is (2kg)*(2 m/s/s), which is 4 Newtons. Now we have enough to solve for F: F = (2kg)*g*cos(60 degrees) - (2kg)*(2 m/s/s) = (2kg)*g*cos(60 degrees) - 4N = (2kg)*g*(0.5) - 4N = (1kg)*g - 4N Now, if you assume that g=9.81m/s/s, then you get: F = 9.81N - 4N = 5.81N Thus, the force pulling the block up the hill is 5.81N. Since it is less than the force pulling it down the hill (which is due to gravity), then the block accelerates down the hill at 2.0 m/s/s. ^okokokko.. where did 60 degrees come from and what does the F stand for???? She totally lost and I hate to ask her to clarify again since I missed it the first and she put so much work into emailing me... can you help?