Posted by Sandhya on Friday, November 7, 2008 at 4:46am.
Kietic friction will apply. There will be a backwards friction force opposing motion that is equal to
M g cos 5 * ěK = 1.95 m/s^2 * M
Motion will stop when the initial kinetic energy equals the gain in potential energy PLUS the work done against friction. Let X be the distance travelled. X sin 5 is the elevation gain.
(1/2) M V^2 = Mg cos5 ěK + MgX sin 5
Cancel out the M's and solve for X
V^2 = 2g [cos5 ěK + sin 5 X]
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