desis Posted October 29, 2008 Posted October 29, 2008 1. Jason takes of across level water on his jet-powered skis. The combined mass of Jason and skis is 75 kg. The skis have a thrust of 200 N and a coeffcient of kinetic friction on water of 0.1. Unfortunately, the skis run out of fuel after 57 s. What is Jason's top speed? I am just doing sample problems for my test today please help thank you
Miri Posted October 29, 2008 Posted October 29, 2008 You have enough information to use the basic kinematics equation vf = vo+at. The force in the forward direction is equal and opposite to the thrust, 200N, minus the force due to friction (F-friction = umg = 0.1*75*9.81). Divide that force by mass to find acceleration, and plug it into the equation. vf is your top speed.
big314mp Posted October 29, 2008 Posted October 29, 2008 http://www.scienceforums.net/forum/showthread.php?t=35529&highlight=Skis There is a fair amount of info here also.
desis Posted October 29, 2008 Author Posted October 29, 2008 Oky I am trying to work this problem out myself but I got stuck I have sollution as well I got till this far 1. F = FT - Mu kmg F = 200 - (0:1)(75)(9:8) = 126 N F = ma = F/m = 126=75 = 1:69 m/s2 s = 1/2at2 = 1/2(1:69)(61)2 = 3144 m<---------- I don't get how they got and where they got 61 from v = at = (1:69)(61) = 103 m/s ]<---------- same for here x = v2/2a; a = Mu kg x = (103)2 2(0:1)(9:8) = 5413 xT = 3144 + 5413 = 8:6 km
Mr Skeptic Posted November 1, 2008 Posted November 1, 2008 You are over-analyzing the problem. All the new information you need is the acceleration, and you already know the time over which acceleration occurs.
swansont Posted November 1, 2008 Posted November 1, 2008 OkyI am trying to work this problem out myself but I got stuck I have sollution as well I got till this far 1. F = FT - Mu kmg F = 200 - (0:1)(75)(9:8) = 126 N F = ma = F/m = 126=75 = 1:69 m/s2 s = 1/2at2 = 1/2(1:69)(61)2 = 3144 m<---------- I don't get how they got and where they got 61 from v = at = (1:69)(61) = 103 m/s ]<---------- same for here x = v2/2a; a = Mu kg x = (103)2 2(0:1)(9:8) = 5413 xT = 3144 + 5413 = 8:6 km Are you sure the solution is to the exact same problem? Sometimes similar problems are written with different values in them.
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