Projectile motion with friction on an inclined plane -- PLEASE HELP!!!!! 1. The problem statement, all variables and given/known data A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 35* angle. The block’s initial speed is 10m/s. (Given that μk = 0.20) a. What vertical height does the block reach above its starting point? b. What speed does it have when it slides back down to its starting point? 2. Relevant equations 3. The attempt at a solution FOR PART A: I know that initial velocity (y component) is Vi = Vi sin theta = 10 sin 35 = 5.73m/s I know we use the equation: Vf^2 = Vi^2 + 2a (deltaY) So we need the value of a . . . Fnet = ma, but we need to knwo the value of Fnet Fnet = Flaunch - Fk? We know that Fk = coeff (m)(g) = 0.2 (2)(-9.8) = -3.92N Do we need to find teh value of the force of the launch? I assumed that the value of the launch can be divided into the x-component and the y-component vi(x) = Vicos theta = 10 cos 35 = 8.19 m/s vi(y) = Vi sin theta = 10 sin 35 = 5.73 m/s and if we plug in those values to get their overall magnitude sqrt ( vi(y)^2 + vi(x)^2 ) sqrt [ (8.19)^2 + (5.73)^2 ] sqrt [ 67.1 + 32.9] sqrt (100) = 10 Flaunch = 10 ( mass) = 10 (2) = 20N Fnet = 20 - 3.92 = 16.08 Fnet = ma 16.08 = (2)a a = 8.04 m/s2 Vf(y)^2 = Vi(y)^2 + 2(ay)(deltaY) 0^2 = 5.73^2 + 2 (-8.04)(deltay) -32.8 = -16.08 (deltaY) deltaY = 2.04m <<< FINAL ANSWER.... is that corect? Everything from the point where I suggested that Fnet = Flaunch - Fk . . . was something I deduced on my own. So I have no idea if it even makes sense.