javagamer Posted August 17, 2010 Share Posted August 17, 2010 (edited) Hi, I have some summer physics work to do online and I've come across what I believe to me a mistake in the site where I do the problems. However, before I send the teacher an email I want to be sure I'm not making some simple mistake. The question in question is #2 at http://wps.prenhall.com/esm_giancoli_physicsppa_6/16/4351/1113975.cw/content/index.html. Note, the numbers involved in the problem change every time (except for the angle in #2 which is always 30 afaik). The equation I've found is: (V^2)/(2 * g (Sin(D) + u)) Where: V is the starting Velocity, g is the (positive) acceleration from Gravity, D is the angle in Degrees, and u is the kinetic friction coefficient. To find this I first drew up a force diagram and found the accelerations from the forces affecting the skier. Gravity: -g * Sin(D) Friction: -g * u I then combined these to get -g * (Sin(D) + u) Then I took the equation F^2 = V^2 + 2 * A * D Where F is the Final velocity and D is the Distance Since I want the furthest the skier traveled I set F to 0 and re-arraged the equation to solve for distance. This gave me: D = V^2/(-2A) I plugged in the accelerations I found to get D = V^2/(2 * g * (Sin(D) + u)) This works for the first question when I have no friction (so u = 0), but I always get one below the answer for part two. So, did I make a mistake or does the script that they use to compute the right answers make a mistake? For those who don't want to click the link here are some sample numbers. 20.5 for the starting velocity, 30 for the angle, and 0.13 for the frictional coefficient. Apparently the right answer is 35 yet I get 34. Edited August 17, 2010 by javagamer Link to comment Share on other sites More sharing options...

Mr Skeptic Posted August 17, 2010 Share Posted August 17, 2010 Your equation looks right, though myself I would have set the kinetic energy equal to the gravitational+friction rather than how you did it. Maybe you and the program have a different value for g. Sounds like the sort of thing your difference between g=10 m/s^2 and g=9.81 m/s^2 would give. Link to comment Share on other sites More sharing options...

javagamer Posted August 17, 2010 Author Share Posted August 17, 2010 (edited) I considered that, but increasing g would only decrease my answer and I'm already one below what it's looking for. However, unless there's some value of g < 9.8 I'm pretty sure that's not it. Turns out I had frictional acceleration wrong because I wasn't taking into account the angle. The acceleration from friction should equal -g * u * Cos(D) which would make my final equation be: distance = V^2/(2 * g * (Sin(D) + uCos(D))) Credit goes to Capn_Refsmmat who helped me on IRC. Edited August 17, 2010 by javagamer Link to comment Share on other sites More sharing options...

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